fizikzauglom

GRADE 10 PRACTICE

Physics Problems: Kinematics, Grade 10

Description of linear and circular motion.

50 problems across 6 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

50 problems
Position in this set: 1 from 50

Problem #1

Kinematics: uniform motion

How long will it take the car to cover the distance? 250\,\text{м} at speed 72\,\text{км/ч}?

Answer options

  1. 6{,}25\,\text{с}
  2. 25\,\text{с}
  3. 125\,\text{с}
  4. 12{,}5\,\text{с}
Show the walkthroughHide the walkthrough
Given
S
250\,\text{м}
v
72\,\text{км/ч}20\,\text{м/с}
Find

t

  1. v = \dfrac{S}{t}

    Uniform motion: speed is the distance divided by the time.

  2. t = \dfrac{S}{v}t = \dfrac{250\,\text{м}}{20\,\tfrac{\text{м}}{\text{с}}}t = 12{,}5\,\text{с}

Answert = 12{,}5\,\text{с}

Position in this set: 2 from 50

Problem #2

Kinematics: uniformly accelerated motion

Having speed 20\,\text{м/с}, the car slows down and stops behind 8\,\text{с}. Determine the acceleration modulus and the distance traveled.

Answer options

  1. \left(2{,}5\,\text{м/с}^{2};\ 80\,\text{м}\right)
  2. \left(1{,}25\,\text{м/с}^{2};\ 40\,\text{м}\right)
  3. \left(5\,\text{м/с}^{2};\ 160\,\text{м}\right)
  4. \left(25\,\text{м/с}^{2};\ 800\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
v_0
20\,\text{м/с}
t
8\,\text{с}
Find

aS

  1. a = \dfrac{v_0}{t}a = \dfrac{20\,\tfrac{\text{м}}{\text{с}}}{8\,\text{с}}a = 2{,}5\,\tfrac{\text{м}}{\text{с}^{2}}

    When braking to a stop, the acceleration module is equal to the initial speed divided by time.

  2. S = \dfrac{v_0}{2}\cdot tS = \dfrac{20\,\tfrac{\text{м}}{\text{с}}}{2}\cdot 8\,\text{с}S = 80\,\text{м}

    The same path is given by writing through acceleration: S = at^2/2 = v_0^2/(2a).

Answera = 2{,}5\,\tfrac{\text{м}}{\text{с}^{2}}S = 80\,\text{м}

Position in this set: 3 from 50

Problem #5

Kinematics: free fall and circular motion

Car wheel radius 30\,\text{см} does it in a minute 300 rpm Determine the period, frequency, angular velocity, linear velocity and centripetal acceleration of the extreme points. Then find the braking distance and acceleration modulus of the car if it stopped for 10\,\text{с}. The wheel rolls without slipping; braking is uniform. Determine the speed and centripetal acceleration of the points of the wheel relative to its axis.

Answer options

  1. \left(0{,}1\,\text{с};\ 2{,}5\,\text{Гц};\ 15{,}7\,\text{рад/с};\ 4{,}71\,\text{м/с};\ 147{,}9\,\text{м/с}^{2};\ 23{,}55\,\text{м};\ 0{,}471\,\text{м/с}^{2}\right)
  2. \left(0{,}4\,\text{с};\ 10\,\text{Гц};\ 62{,}8\,\text{рад/с};\ 18{,}84\,\text{м/с};\ 591{,}6\,\text{м/с}^{2};\ 94{,}2\,\text{м};\ 1{,}884\,\text{м/с}^{2}\right)
  3. \left(2\,\text{с};\ 50\,\text{Гц};\ 314\,\text{рад/с};\ 94{,}2\,\text{м/с};\ 2958\,\text{м/с}^{2};\ 471\,\text{м};\ 9{,}42\,\text{м/с}^{2}\right)
  4. \left(0{,}2\,\text{с};\ 5\,\text{Гц};\ 31{,}4\,\text{рад/с};\ 9{,}42\,\text{м/с};\ 295{,}8\,\text{м/с}^{2};\ 47{,}1\,\text{м};\ 0{,}942\,\text{м/с}^{2}\right)
Show the walkthroughHide the walkthrough
Given
R
30\,\text{см}0{,}3\,\text{м}
t_{\text{об}}
1\,\text{мин}60\,\text{с}
n
300
t
10\,\text{с}
Find

T\nu\omegava_{\text{цс}}Sa

  1. T = \dfrac{t_{\text{об}}}{n}T = \dfrac{60\,\text{с}}{300}T = 0{,}2\,\text{с}
  2. \nu = \dfrac{n}{t_{\text{об}}} = \dfrac{1}{T}\nu = \dfrac{300}{60\,\text{с}}\nu = 5\,\text{Гц}
  3. \omega = 2\pi\nu\omega = 2\cdot 3{,}14\cdot 5\,\text{Гц}\omega = 31{,}4\,\tfrac{\text{рад}}{\text{с}}
  4. v = \dfrac{2\pi R}{T} = 2\pi R\nuv = 2\cdot 3{,}14\cdot 0{,}3\,\text{м}\cdot 5\,\text{Гц}v = 9{,}42\,\tfrac{\text{м}}{\text{с}}
  5. a_{\text{цс}} = \dfrac{v^{2}}{R}a_{\text{цс}} = \dfrac{9{,}42^{2}\,\tfrac{\text{м}^{2}}{\text{с}^{2}}}{0{,}3\,\text{м}}a_{\text{цс}} = 295{,}8\,\tfrac{\text{м}}{\text{с}^{2}}
  6. S = \dfrac{v}{2}\cdot tS = \dfrac{9{,}42\,\tfrac{\text{м}}{\text{с}}}{2}\cdot 10\,\text{с}S = 47{,}1\,\text{м}

    When moving equally slow to a stop, the average speed is half the initial speed.

  7. a = \dfrac{v}{t}a = \dfrac{9{,}42\,\tfrac{\text{м}}{\text{с}}}{10\,\text{с}}a = 0{,}942\,\tfrac{\text{м}}{\text{с}^{2}}

AnswerT = 0{,}2\,\text{с}\nu = 5\,\text{Гц}\omega = 31{,}4\,\tfrac{\text{рад}}{\text{с}}v = 9{,}42\,\tfrac{\text{м}}{\text{с}}a_{\text{цс}} = 295{,}8\,\tfrac{\text{м}}{\text{с}^{2}}S = 47{,}1\,\text{м}a = 0{,}942\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 4 from 50

Problem #6

Kinematics: free fall and circular motion

A body is thrown vertically upward with a speed 40\,\text{м/с}. How long will it take for it to return, what maximum height will it reach and what height will it be at in 2\,\text{с}?

Answer options

  1. \left(8\,\text{с};\ 80\,\text{м};\ 60\,\text{м}\right)
  2. \left(4\,\text{с};\ 40\,\text{м};\ 30\,\text{м}\right)
  3. \left(16\,\text{с};\ 160\,\text{м};\ 120\,\text{м}\right)
  4. \left(80\,\text{с};\ 800\,\text{м};\ 600\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
v_0
40\,\text{м/с}
t_h
2\,\text{с}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

tHh

  1. H = \dfrac{v_0^{2}}{2g}H = \dfrac{40^{2}\,\tfrac{\text{м}^{2}}{\text{с}^{2}}}{2\cdot 10\,\tfrac{\text{м}}{\text{с}^{2}}}H = 80\,\text{м}
  2. t_1 = \dfrac{v_0}{g}t_1 = \dfrac{40\,\tfrac{\text{м}}{\text{с}}}{10\,\tfrac{\text{м}}{\text{с}^{2}}}t_1 = 4\,\text{с}

    Ascent and descent take the same time, so t_2 = t_1.

  3. t = t_1 + t_2 = 2t_1t = 2\cdot 4\,\text{с}t = 8\,\text{с}
  4. h = v_0 t_h - \dfrac{g t_h^{2}}{2}h = 40\,\tfrac{\text{м}}{\text{с}}\cdot 2\,\text{с} - \dfrac{10\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 2^{2}\,\text{с}^{2}}{2}h = 80\,\text{м} - 20\,\text{м} = 60\,\text{м}

Answert = 8\,\text{с}H = 80\,\text{м}h = 60\,\text{м}

Position in this set: 5 from 50

Problem #7

Kinematics: uniform motion

The helicopter flew north 400\,\text{км}, turned to 90^{\circ} and flew by 300\,\text{км} to the east. Find the path and the moving module.

Answer options

  1. \left(7000\,\text{км};\ 5000\,\text{км}\right)
  2. \left(700\,\text{км};\ 500\,\text{км}\right)
  3. \left(350\,\text{км};\ 250\,\text{км}\right)
  4. \left(1400\,\text{км};\ 1000\,\text{км}\right)
Show the walkthroughHide the walkthrough
Given
S_1
400\,\text{км}
S_2
300\,\text{км}
Find

\ellS

  1. \ell = S_1 + S_2\ell = 400\,\text{км} + 300\,\text{км}\ell = 700\,\text{км}

    The path is the sum of the lengths of both sections.

  2. S = \sqrt{S_1^{2} + S_2^{2}}S = \sqrt{400^{2} + 300^{2}}\,\text{км}S = 500\,\text{км}

    The sections are perpendicular, so the displacement is the hypotenuse.

Answer\ell = 700\,\text{км}S = 500\,\text{км}

Position in this set: 6 from 50

Problem #8

Kinematics: uniform motion

A freight train left the station at a speed 60\,\text{км/ч}. Half an hour later, an electric train left in the same direction at a speed 90\,\text{км/ч}. How long after the train leaves and at what distance from the station will it catch up with the train?

Answer options

  1. \left(2\,\text{ч};\ 180\,\text{км}\right)
  2. \left(10\,\text{ч};\ 900\,\text{км}\right)
  3. \left(1\,\text{ч};\ 90\,\text{км}\right)
  4. \left(0{,}5\,\text{ч};\ 45\,\text{км}\right)
Show the walkthroughHide the walkthrough
Given
v_1
60\,\text{км/ч}
\Delta t
0{,}5\,\text{ч}
v_2
90\,\text{км/ч}
Find

tS

  1. x_1=v_1(t+\Delta t),\quad x_2=v_2t

    Time t is counted from the departure of the train. The freight train had already been traveling for half an hour at this point.

  2. x_1=x_2\ \Rightarrow\ t=\dfrac{v_1\Delta t}{v_2-v_1}t = \dfrac{60\cdot0{,}5}{90-60}t = 1\,\text{ч}
  3. S=v_2tS = 90\cdot1S = 90\,\text{км}

Answert=1\,\text{ч}S=90\,\text{км}

Position in this set: 7 from 50

Problem #9

Kinematics: uniformly accelerated motion

The car slows down with acceleration 4\,\text{м/с}^{2} and stops behind 6\,\text{с}. Determine the braking distance and initial speed. Did the driver violate the restriction? 90\,\text{км/ч}?

Answer options

  1. \left(48\,\text{м};\ 16\,\text{м/с};\ \text{нет}\right)
  2. \left(72\,\text{м};\ 24\,\text{м/с};\ \text{да}\right)
  3. \left(144\,\text{м};\ 48\,\text{м/с};\ \text{да}\right)
  4. \left(72\,\text{м};\ 24\,\text{м/с};\ \text{нет}\right)
Show the walkthroughHide the walkthrough
Given
a
4\,\tfrac{\text{м}}{\text{с}^{2}}
t
6\,\text{с}
v
0
Find

Sv_0\text{conclusion}

  1. S = \dfrac{at^{2}}{2}S = \dfrac{4\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 6^{2}\,\text{с}^{2}}{2}S = 72\,\text{м}
  2. v_0 = atv_0 = 4\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 6\,\text{с}v_0 = 24\,\tfrac{\text{м}}{\text{с}} = 86{,}4\,\tfrac{\text{км}}{\text{ч}}
  3. 86{,}4\,\tfrac{\text{км}}{\text{ч}} < 90\,\tfrac{\text{км}}{\text{ч}}\text{limitation not violated}

AnswerS = 72\,\text{м}v_0 = 24\,\tfrac{\text{м}}{\text{с}}\text{not violated}

Position in this set: 8 from 50

Problem #10

Kinematics: free fall and circular motion

Body thrown with speed 40\,\text{м/с}. It is required to determine the movement for a throw from a height 45\,\text{м} and from the surface of the Earth, but the direction of the initial velocity is not specified. Is there enough initial data for a definite answer?

Answer options

  1. \text{No, direction speed not given}
  2. \text{Yes, enough only module speed}
  3. \text{No, not given Object mass}
  4. \text{Yes, if accept mass equal 1 kg}
Position in this set: 9 from 50

Problem #11

Kinematics: free fall and circular motion

A body falls freely from a height 125\,\text{м}. After how long and at what speed will it fall to Earth? Accept g=10\,\text{м/с}^{2}.

Answer options

  1. \left(50\,\text{с};\ 500\,\text{м/с}\right)
  2. \left(5\,\text{с};\ 50\,\text{м/с}\right)
  3. \left(2{,}5\,\text{с};\ 25\,\text{м/с}\right)
  4. \left(10\,\text{с};\ 100\,\text{м/с}\right)
Show the walkthroughHide the walkthrough
Given
h
125\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

tv

  1. h = \dfrac{gt^{2}}{2}\ \Rightarrow\ t = \sqrt{\dfrac{2h}{g}}t = \sqrt{\dfrac{2\cdot 125\,\text{м}}{10\,\tfrac{\text{м}}{\text{с}^{2}}}}t = 5\,\text{с}
  2. v = gtv = 10\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 5\,\text{с}v = 50\,\tfrac{\text{м}}{\text{с}}

    The same result is given by the formula v = sqrt{2gh}.

Answert = 5\,\text{с}v = 50\,\tfrac{\text{м}}{\text{с}}

Position in this set: 10 from 50

Problem #12

Kinematics: free fall and circular motion

A body is thrown upward with speed 30\,\text{м/с}. How long will it take for it to be up? 40\,\text{м}? Will it reach the heights 50\,\text{м}?

Answer options

  1. \left(t=3\,\text{с};\ \text{50 м недостижимы}\right)
  2. \left(t=2\,\text{с}\ \text{или}\ 4\,\text{с};\ \text{50 м достижимы}\right)
  3. \left(t=2\,\text{с}\ \text{или}\ 4\,\text{с};\ \text{50 м недостижимы}\right)
  4. \left(t=1\,\text{с}\ \text{или}\ 5\,\text{с};\ \text{50 м достижимы}\right)
Show the walkthroughHide the walkthrough
Given
v_0
30\,\text{м/с}
h_1
40\,\text{м}
h_2
50\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

th_2 = 50\,\text{м}

  1. h = v_0 t - \dfrac{gt^{2}}{2}40 = 30t - \dfrac{10t^{2}}{2}5t^{2} - 30t + 40 = 0
  2. t^{2} - 6t + 8 = 0t_{1,2} = \dfrac{6 \pm \sqrt{36 - 32}}{2} = \dfrac{6 \pm 2}{2}t_1 = 2\,\text{с},\quad t_2 = 4\,\text{с}

    Two roots: the body passes this height on the ascent and on the descent.

  3. 50 = 30t - \dfrac{10t^{2}}{2}t^{2} - 6t + 10 = 0,\quad D = 36 - 40 = -4D < 0

    The discriminant is negative, which means the body does not reach such a height.

Answert = 2\,\text{с}\ \text{или}\ 4\,\text{с}\text{высота }50\,\text{м не достигается}

Position in this set: 11 from 50

Problem #13

Kinematics: free fall and circular motion

A body is thrown from the Earth at an angle 60^{\circ} towards the horizon at speed 40\,\text{м/с}. Determine the flight range and maximum lift height.

Answer options

  1. \left(69{,}3\,\text{м};\ 30\,\text{м}\right)
  2. \left(277{,}2\,\text{м};\ 120\,\text{м}\right)
  3. \left(1386\,\text{м};\ 600\,\text{м}\right)
  4. \left(138{,}6\,\text{м};\ 60\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
\alpha
60^{\circ}
v_0
40\,\text{м/с}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

SH

  1. v_{0x} = v_0\cos\alphav_{0x} = 40\,\tfrac{\text{м}}{\text{с}}\cdot 0{,}5v_{0x} = 20\,\tfrac{\text{м}}{\text{с}}
  2. v_{0y} = v_0\sin\alphav_{0y} = 40\,\tfrac{\text{м}}{\text{с}}\cdot 0{,}866v_{0y} = 34{,}64\,\tfrac{\text{м}}{\text{с}}
  3. H = \dfrac{v_{0y}^{2}}{2g}H = \dfrac{34{,}64^{2}}{2\cdot 10}\,\text{м}H = 60\,\text{м}
  4. t_1 = \dfrac{v_{0y}}{g}t_1 = \dfrac{34{,}64}{10}\,\text{с}t_1 = 3{,}464\,\text{с}
  5. t = 2t_1t = 2\cdot 3{,}464\,\text{с}t = 6{,}928\,\text{с}
  6. S = v_{0x}\,tS = 20\,\tfrac{\text{м}}{\text{с}}\cdot 6{,}928\,\text{с}S = 138{,}6\,\text{м}

AnswerS = 138{,}6\,\text{м}H = 60\,\text{м}

Position in this set: 12 from 50

Problem #14

Kinematics: free fall and circular motion

Body thrown from a height 45\,\text{м} at an angle 30^{\circ} towards the horizon at speed 40\,\text{м/с}. At what speed will it fall to Earth?

Answer options

  1. 50\,\text{м/с}
  2. 25\,\text{м/с}
  3. 100\,\text{м/с}
  4. 500\,\text{м/с}
Position in this set: 13 from 50

Problem #15

Kinematics: free fall and circular motion

The balloon begins to rise from the Earth from rest with acceleration 2\,\text{м/с}^{2}. Via 5\,\text{с} a stone fell out of it. How long after separation and at what speed will the stone fall to Earth?

Answer options

  1. \left(34{,}5\,\text{с};\ 245\,\text{м/с}\right)
  2. \left(3{,}45\,\text{с};\ 24{,}5\,\text{м/с}\right)
  3. \left(1{,}725\,\text{с};\ 12{,}25\,\text{м/с}\right)
  4. \left(6{,}9\,\text{с};\ 49\,\text{м/с}\right)
Position in this set: 14 from 50

Problem #16

Kinematics: free fall and circular motion

A body is thrown horizontally with a speed 15\,\text{м/с} and fell to Earth through 2\,\text{с}. Determine the height, flight distance and speed when falling.

Answer options

  1. \left(40\,\text{м};\ 60\,\text{м};\ 50\,\text{м/с}\right)
  2. \left(200\,\text{м};\ 300\,\text{м};\ 250\,\text{м/с}\right)
  3. \left(20\,\text{м};\ 30\,\text{м};\ 25\,\text{м/с}\right)
  4. \left(10\,\text{м};\ 15\,\text{м};\ 12{,}5\,\text{м/с}\right)
Show the walkthroughHide the walkthrough
Given
v_0
15\,\text{м/с}
t
2\,\text{с}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

hSv

  1. h = \dfrac{gt^{2}}{2}h = \dfrac{10\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 2^{2}\,\text{с}^{2}}{2}h = 20\,\text{м}

    With a horizontal throw, the initial vertical velocity is zero.

  2. S = v_0 tS = 15\,\tfrac{\text{м}}{\text{с}}\cdot 2\,\text{с}S = 30\,\text{м}
  3. v_x = v_0,\quad v_y = gtv_y = 10\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 2\,\text{с}v_y = 20\,\tfrac{\text{м}}{\text{с}}
  4. v = \sqrt{v_x^{2} + v_y^{2}}\sqrt{15^{2} + 20^{2}} = \sqrt{625}v = 25\,\tfrac{\text{м}}{\text{с}}

Answerh = 20\,\text{м}S = 30\,\text{м}v = 25\,\tfrac{\text{м}}{\text{с}}

Position in this set: 15 from 50

Problem #17

Kinematics: free fall and circular motion

Body thrown with speed 30\,\text{м/с} at an angle to the horizontal. The maximum lift height is 26{,}5\,\text{м}. Determine the throwing angle and flight distance.

Answer options

  1. \left(25\,^{\circ};\ 44{,}3\,\text{м}\right)
  2. \left(100\,^{\circ};\ 177{,}2\,\text{м}\right)
  3. \left(500\,^{\circ};\ 886\,\text{м}\right)
  4. \left(50\,^{\circ};\ 88{,}6\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
v_0
30\,\text{м/с}
H
26{,}5\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

\alphaS

  1. H = \dfrac{v_{0y}^{2}}{2g}\ \Rightarrow\ v_{0y} = \sqrt{2gH}v_{0y} = \sqrt{2\cdot 10\cdot 26{,}5}v_{0y} = 23\,\tfrac{\text{м}}{\text{с}}
  2. \sin\alpha = \dfrac{v_{0y}}{v_0}\sin\alpha = \dfrac{23}{30}\alpha = 0{,}7667\ \Rightarrow\ \alpha = 50^{\circ}
  3. v_{0x} = v_0\cos\alphav_{0x} = 30\cdot 0{,}642v_{0x} = 19{,}3\,\tfrac{\text{м}}{\text{с}}
  4. t = \dfrac{2v_{0y}}{g}t = \dfrac{2\cdot 23}{10}t = 4{,}6\,\text{с}
  5. S = v_{0x}\,tS = 19{,}3\cdot 4{,}6S = 88{,}6\,\text{м}

Answer\alpha = 50^{\circ}S = 88{,}6\,\text{м}

Position in this set: 16 from 50

Problem #18

Kinematics: free fall and circular motion

The car is moving at speed 90\,\text{км/ч}. Wheel diameter 60\,\text{см}. Determine the period, frequency, angular velocity and centripetal acceleration of the extreme points of the wheel. The car stops for 5\,\text{с}: Find the braking distance and the number of revolutions during braking. The wheel rolls without slipping, braking is uniformly slow. Determine the rotation characteristics relative to the wheel axis.

Answer options

  1. \left(0{,}0754\,\text{с};\ 13{,}26\,\text{Гц};\ 83{,}33\,\text{рад/с};\ 2083\,\text{м/с}^{2};\ 62{,}5\,\text{м};\ 33{,}2\,\text{оборотов}\right)
  2. \left(0{,}0377\,\text{с};\ 6{,}63\,\text{Гц};\ 41{,}665\,\text{рад/с};\ 1041{,}5\,\text{м/с}^{2};\ 31{,}25\,\text{м};\ 16{,}6\,\text{оборотов}\right)
  3. \left(0{,}1508\,\text{с};\ 26{,}52\,\text{Гц};\ 166{,}66\,\text{рад/с};\ 4166\,\text{м/с}^{2};\ 125\,\text{м};\ 66{,}4\,\text{оборотов}\right)
  4. \left(0{,}754\,\text{с};\ 132{,}6\,\text{Гц};\ 833{,}3\,\text{рад/с};\ 20830\,\text{м/с}^{2};\ 625\,\text{м};\ 332\,\text{оборотов}\right)
Position in this set: 17 from 50

Problem #19

Kinematics: free fall and circular motion

From what height did the body fall if during the last second of the fall it flew: a) 25\,\text{м}, b) 40\,\text{м}? The body is released without an initial velocity; air resistance is neglected.

Answer options

  1. \left(450\,\text{м};\ 1012{,}5\,\text{м}\right)
  2. \left(45\,\text{м};\ 101{,}25\,\text{м}\right)
  3. \left(22{,}5\,\text{м};\ 50{,}625\,\text{м}\right)
  4. \left(90\,\text{м};\ 202{,}5\,\text{м}\right)
Position in this set: 18 from 50

Problem #20

Kinematics: uniform motion

The tourist passed 5\,\text{км} north, then some more 12\,\text{км} in an unknown direction. Determine the path and possible range of the motion module.

Answer options

  1. \left(17\,\text{км};\ |\Delta r|=17\,\text{км}\right)
  2. \left(7\,\text{км};\ |\Delta r|=12\,\text{км}\right)
  3. \left(17\,\text{км};\ 7\,\text{км}\leq |\Delta r|\leq17\,\text{км}\right)
  4. \left(12\,\text{км};\ 5\,\text{км}\leq |\Delta r|\leq12\,\text{км}\right)
Position in this set: 19 from 50

Problem #21

Kinematics: uniform motion

Distance between piers 60\,\text{км}. Current speed 5\,\text{км/ч}, speed of the boat relative to the water 15\,\text{км/ч}. How long will it take to get there and back?

Answer options

  1. 4{,}5\,\text{ч}
  2. 18\,\text{ч}
  3. 90\,\text{ч}
  4. 9\,\text{ч}
Show the walkthroughHide the walkthrough
Given
S
60\,\text{км}
v_{\text{т}}
5\,\text{км/ч}
v_{\text{к}}
15\,\text{км/ч}
Find

t

  1. t = \dfrac{S}{v_{\text{к}} + v_{\text{т}}} + \dfrac{S}{v_{\text{к}} - v_{\text{т}}}

    Velocities along the current are added, while against the current they are subtracted.

  2. t = \dfrac{60}{15 + 5} + \dfrac{60}{15 - 5}t = 3\,\text{ч} + 6\,\text{ч}t = 9\,\text{ч}

Answert = 9\,\text{ч}

Position in this set: 20 from 50

Problem #22

Kinematics: free fall and circular motion

The car moves around a bend with a radius 160\,\text{м}. What is its speed if the centripetal acceleration is 2{,}5\,\text{м/с}^{2}?

Answer options

  1. 20\,\text{м/с}
  2. 10\,\text{м/с}
  3. 40\,\text{м/с}
  4. 200\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
R
160\,\text{м}
a_{\text{цс}}
2{,}5\,\tfrac{\text{м}}{\text{с}^{2}}
Find

v

  1. a_{\text{цс}} = \dfrac{v^{2}}{R}\ \Rightarrow\ v = \sqrt{a_{\text{цс}}R}\sqrt{2{,}5\,\tfrac{\text{м}}{\text{с}^{2}}\cdot 160\,\text{м}} = \sqrt{400\,\tfrac{\text{м}^{2}}{\text{с}^{2}}}v = 20\,\tfrac{\text{м}}{\text{с}}

Answerv = 20\,\tfrac{\text{м}}{\text{с}}

Position in this set: 21 from 50

Problem #23

Kinematics: uniformly accelerated motion

The body moves in a straight line without changing direction with an initial speed 10\,\text{м/с} and constant acceleration. For 5\,\text{с} it passes 40\,\text{м}. Find the projection of acceleration onto the axis directed along the initial velocity.

Answer options

  1. -8\,\text{м/с}^{2}
  2. -0{,}8\,\text{м/с}^{2}
  3. -0{,}4\,\text{м/с}^{2}
  4. -1{,}6\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
v_0
10\,\text{м/с}
t
5\,\text{с}
S
40\,\text{м}
Find

a_x

  1. S = v_0 t + \dfrac{a_x t^{2}}{2}\ \Rightarrow\ 2S = 2v_0 t + a_x t^{2}

    Multiply by two to get rid of the fraction.

  2. a_x = \dfrac{2S - 2v_0 t}{t^{2}}\dfrac{2\cdot 40 - 2\cdot 10\cdot 5}{5^{2}} = \dfrac{80 - 100}{25}a_x = -0{,}8\,\tfrac{\text{м}}{\text{с}^{2}}

    The projection is negative: acceleration is directed against the initial velocity. At the end of the interval, the speed is 6 m/s, the direction of movement does not change.

Answera_x=-0{,}8\,\text{м/с}^{2}

Position in this set: 22 from 50

Problem #24

Kinematics: free fall and circular motion

The runner moves in a circle with a radius 20\,\text{м}. What are the path and the displacement module after 1{,}5 circle?

Answer options

  1. \left(377\,\text{м};\ 80\,\text{м}\right)
  2. \left(1885\,\text{м};\ 400\,\text{м}\right)
  3. \left(188{,}5\,\text{м};\ 40\,\text{м}\right)
  4. \left(94{,}25\,\text{м};\ 20\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
R
20\,\text{м}
n
1{,}5
Find

\ellS

  1. \ell = n\cdot 2\pi R\ell = 1{,}5\cdot 2\cdot 3{,}14\cdot 20\,\text{м}\ell = 188{,}5\,\text{м}
  2. S = 2RS = 2\cdot 20\,\text{м}S = 40\,\text{м}

    After one and a half laps, the runner finds himself at a diametrically opposite point.

Answer\ell = 188{,}5\,\text{м}S = 40\,\text{м}

Position in this set: 23 from 50

Problem #25

Kinematics: uniform motion

The pedestrian passed 300\,\text{м} at speed 1{,}5\,\text{м/с}, then again 2\,\text{мин} moved at speed 2{,}5\,\text{м/с}. Determine the average speed along the entire path.

Answer options

  1. 0{,}9375\,\text{м/с}
  2. 3{,}75\,\text{м/с}
  3. 18{,}75\,\text{м/с}
  4. 1{,}875\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
S_1
300\,\text{м}
v_1
1{,}5\,\text{м/с}
t_2
2\,\text{мин}120\,\text{с}
v_2
2{,}5\,\text{м/с}
Find

v_{\text{ср}}

  1. t_1 = \dfrac{S_1}{v_1}t_1 = \dfrac{300}{1{,}5}t_1 = 200\,\text{с}
  2. S_2 = v_2 t_2S_2 = 2{,}5\cdot 120S_2 = 300\,\text{м}
  3. v_{\text{ср}} = \dfrac{S_1 + S_2}{t_1 + t_2}\dfrac{300 + 300}{200 + 120} = \dfrac{600}{320}v_{\text{ср}} = 1{,}875\,\tfrac{\text{м}}{\text{с}}

    Average speed is the entire path divided by the entire time, not the average of speeds.

Answerv_{\text{ср}} = 1{,}875\,\tfrac{\text{м}}{\text{с}}

Position in this set: 24 from 50

Problem #26

Kinematics: free fall and circular motion

Body thrown at an angle 60^{\circ} to the horizon and rose to a height 34\,\text{м}. Determine the initial speed and flight range.

Answer options

  1. \left(30\,\text{м/с};\ 78\,\text{м}\right)
  2. \left(15\,\text{м/с};\ 39\,\text{м}\right)
  3. \left(60\,\text{м/с};\ 156\,\text{м}\right)
  4. \left(300\,\text{м/с};\ 780\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
\alpha
60^{\circ}
H
34\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

v_0S

  1. H = \dfrac{v_{0y}^{2}}{2g}\ \Rightarrow\ v_{0y} = \sqrt{2gH}v_{0y} = \sqrt{2\cdot 10\cdot 34}v_{0y} = 26\,\tfrac{\text{м}}{\text{с}}

    The lifting height depends only on the vertical component of the speed.

  2. v_{0y} = v_0\sin\alpha\ \Rightarrow\ v_0 = \dfrac{v_{0y}}{\sin\alpha}\dfrac{26}{\sin 60^{\circ}} = \dfrac{26}{0{,}866}v_0 = 30\,\tfrac{\text{м}}{\text{с}}
  3. v_{0x} = v_0\cos\alphav_{0x} = 30\cdot 0{,}5v_{0x} = 15\,\tfrac{\text{м}}{\text{с}}
  4. t = \dfrac{2v_{0y}}{g}t = \dfrac{2\cdot 26}{10}t = 5{,}2\,\text{с}
  5. S = v_{0x}\,tS = 15\cdot 5{,}2S = 78\,\text{м}

Answerv_0 = 30\,\tfrac{\text{м}}{\text{с}}S = 78\,\text{м}

Position in this set: 25 from 50

Problem #27

Kinematics: uniform motion

How far will the plane fly in 15\,\text{мин}, if its speed is 900\,\text{км/ч}?

Answer options

  1. 2250\,\text{км}
  2. 225\,\text{км}
  3. 112{,}5\,\text{км}
  4. 450\,\text{км}
Show the walkthroughHide the walkthrough
Given
t
15\,\text{мин}900\,\text{с}
v
900\,\text{км/ч}250\,\text{м/с}
Find

S

  1. S = vtS = 250\,\tfrac{\text{м}}{\text{с}}\cdot 900\,\text{с}S = 225\,000\,\text{м} = 225\,\text{км}

AnswerS = 225\,\text{км}

Position in this set: 26 from 50

Problem #28

Kinematics: free fall and circular motion

A body is thrown horizontally from a height 20\,\text{м} and flew horizontally 30\,\text{м}. Determine the speed when falling to Earth.

Answer options

  1. 50\,\text{м/с}
  2. 250\,\text{м/с}
  3. 25\,\text{м/с}
  4. 12{,}5\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
h
20\,\text{м}
S
30\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

v

  1. h = \dfrac{gt^{2}}{2}\ \Rightarrow\ t = \sqrt{\dfrac{2h}{g}}t = \sqrt{\dfrac{2\cdot 20}{10}}t = 2\,\text{с}

    The initial vertical velocity during a horizontal throw is zero.

  2. v_y = gtv_y = 10\cdot 2v_y = 20\,\tfrac{\text{м}}{\text{с}}
  3. v_x = v_{0x} = \dfrac{S}{t}v_x = \dfrac{30}{2}v_x = 15\,\tfrac{\text{м}}{\text{с}}
  4. v = \sqrt{v_x^{2} + v_y^{2}}v = \sqrt{15^{2} + 20^{2}}v = 25\,\tfrac{\text{м}}{\text{с}}

Answerv = 25\,\tfrac{\text{м}}{\text{с}}

Position in this set: 27 from 50

Problem #29

Kinematics: uniformly accelerated motion

On the way 30\,\text{м} the car changed speed from 72\,\text{км/ч} up to 36\,\text{км/ч}. Determine the acceleration.

Answer options

  1. -2{,}5\,\text{м/с}^{2}
  2. -10\,\text{м/с}^{2}
  3. -50\,\text{м/с}^{2}
  4. -5\,\text{м/с}^{2}
Show the walkthroughHide the walkthrough
Given
S
30\,\text{м}
v_0
72\,\text{км/ч}20\,\text{м/с}
v
36\,\text{км/ч}10\,\text{м/с}
Find

a_x

  1. S = \dfrac{v^{2} - v_0^{2}}{2a_x}\ \Rightarrow\ a_x = \dfrac{v^{2} - v_0^{2}}{2S}

    The formula without time connects speeds and paths directly.

  2. a_x = \dfrac{v^{2} - v_0^{2}}{2S}\dfrac{10^{2} - 20^{2}}{2\cdot 30} = \dfrac{100 - 400}{60}a_x = -5\,\tfrac{\text{м}}{\text{с}^{2}}

    The minus sign means braking; acceleration module is 5 m/s².

Answera_x = -5\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 28 from 50

Problem #30

Kinematics: uniformly accelerated motion

In uniformly accelerated motion with acceleration 1{,}5\,\text{м/с}^{2} the body has passed 87\,\text{м} for 6\,\text{с}. What is the initial speed?

Answer options

  1. 10\,\text{м/с}
  2. 5\,\text{м/с}
  3. 20\,\text{м/с}
  4. 100\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
a
1{,}5\,\tfrac{\text{м}}{\text{с}^{2}}
S
87\,\text{м}
t
6\,\text{с}
Find

v_0

  1. S = v_0 t \pm \dfrac{at^{2}}{2}\ \Rightarrow\ v_0 = \dfrac{2S \mp at^{2}}{2t}

    The sign depends on whether the body is accelerating or decelerating.

  2. v_{0} = \dfrac{2S - at^{2}}{2t}\dfrac{2\cdot 87 - 1{,}5\cdot 6^{2}}{2\cdot 6} = \dfrac{174 - 54}{12}v_0 = 10\,\tfrac{\text{м}}{\text{с}}

    Acceleration: acceleration is directed along the movement - this case is implied in the condition.

  3. v_{0} = \dfrac{2S + at^{2}}{2t}v_{0} = \dfrac{174 + 54}{12}v_{0} = 19\,\tfrac{\text{м}}{\text{с}}

    For comparison: this would be the case when braking with the same acceleration module.

Answerv_0 = 10\,\tfrac{\text{м}}{\text{с}}

Position in this set: 29 from 50

Problem #31

Kinematics: free fall and circular motion

Body thrown at an angle 45^{\circ} towards the horizon, flew horizontally 22{,}5\,\text{м}. Determine the initial speed and maximum altitude.

Answer options

  1. \left(150\,\text{м/с};\ 56{,}25\,\text{м}\right)
  2. \left(15\,\text{м/с};\ 5{,}625\,\text{м}\right)
  3. \left(7{,}5\,\text{м/с};\ 2{,}8125\,\text{м}\right)
  4. \left(30\,\text{м/с};\ 11{,}25\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
\alpha
45^{\circ}
S
22{,}5\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

v_0H

  1. S = v_{0x}t = v_{0x}\cdot\dfrac{2v_{0y}}{g} = \dfrac{2v_{0y}^{2}}{g}

    At an angle of 45°, the horizontal and vertical components are equal.

  2. v_{0y} = \sqrt{\dfrac{Sg}{2}}\sqrt{\dfrac{22{,}5\cdot 10}{2}} = \sqrt{112{,}5}v_{0y} = 10{,}6\,\tfrac{\text{м}}{\text{с}}
  3. v_0 = \dfrac{v_{0y}}{\sin\alpha}v_0 = \dfrac{10{,}6}{\sin 45^{\circ}}v_0 = 15\,\tfrac{\text{м}}{\text{с}}
  4. H = \dfrac{v_{0y}^{2}}{2g}H = \dfrac{112{,}5}{2\cdot 10}H = 5{,}625\,\text{м}

Answerv_0 = 15\,\tfrac{\text{м}}{\text{с}}H = 5{,}625\,\text{м}

Position in this set: 30 from 50

Problem #32

Kinematics: free fall and circular motion

A body is thrown horizontally and falls through 4\,\text{с}, flying horizontally 120\,\text{м}. Determine the height, initial horizontal velocity, and falling velocity.

Answer options

  1. \left(160\,\text{м};\ 60\,\text{м/с};\ 100\,\text{м/с}\right)
  2. \left(800\,\text{м};\ 300\,\text{м/с};\ 500\,\text{м/с}\right)
  3. \left(80\,\text{м};\ 30\,\text{м/с};\ 50\,\text{м/с}\right)
  4. \left(40\,\text{м};\ 15\,\text{м/с};\ 25\,\text{м/с}\right)
Show the walkthroughHide the walkthrough
Given
t
4\,\text{с}
S
120\,\text{м}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

hv_{0x}v

  1. h = \dfrac{gt^{2}}{2}h = \dfrac{10\cdot 4^{2}}{2}h = 80\,\text{м}
  2. v_x = v_{0x} = \dfrac{S}{t}v_x = \dfrac{120}{4}v_{0x} = 30\,\tfrac{\text{м}}{\text{с}}
  3. v_y = gtv_y = 10\cdot 4v_y = 40\,\tfrac{\text{м}}{\text{с}}
  4. v = \sqrt{v_x^{2} + v_y^{2}}v = \sqrt{30^{2} + 40^{2}}v = 50\,\tfrac{\text{м}}{\text{с}}

Answerh = 80\,\text{м}v_{0x} = 30\,\tfrac{\text{м}}{\text{с}}v = 50\,\tfrac{\text{м}}{\text{с}}

Position in this set: 31 from 50

Problem #33

Kinematics: free fall and circular motion

Body thrown at an angle 45^{\circ} to the horizon, fell to Earth through 5\,\text{с}. Determine the initial speed, flight range and maximum altitude.

Answer options

  1. \left(17{,}68\,\text{м/с};\ 62{,}5\,\text{м};\ 15{,}625\,\text{м}\right)
  2. \left(70{,}72\,\text{м/с};\ 250\,\text{м};\ 62{,}5\,\text{м}\right)
  3. \left(353{,}6\,\text{м/с};\ 1250\,\text{м};\ 312{,}5\,\text{м}\right)
  4. \left(35{,}36\,\text{м/с};\ 125\,\text{м};\ 31{,}25\,\text{м}\right)
Show the walkthroughHide the walkthrough
Given
\alpha
45^{\circ}
t
5\,\text{с}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

v_0SH

  1. t = 2t_1\ \Rightarrow\ t_1 = \dfrac{t}{2}t_1 = \dfrac{5}{2}t_1 = 2{,}5\,\text{с}
  2. v_{0y} = g t_1 = v_{0x}v_{0y} = 10\cdot 2{,}5v_{0y} = 25\,\tfrac{\text{м}}{\text{с}}

    At 45° the components are equal.

  3. S = v_{0x}\,tS = 25\cdot 5S = 125\,\text{м}
  4. v_0 = \dfrac{v_{0y}}{\sin\alpha}v_0 = \dfrac{25}{\sin 45^{\circ}}v_0 = 35{,}36\,\tfrac{\text{м}}{\text{с}}
  5. H = \dfrac{v_{0y}^{2}}{2g}H = \dfrac{25^{2}}{2\cdot 10}H = 31{,}25\,\text{м}

Answerv_0 = 35{,}36\,\tfrac{\text{м}}{\text{с}}S = 125\,\text{м}H = 31{,}25\,\text{м}

Position in this set: 32 from 50

Problem #34

Kinematics: uniformly accelerated motion

The body has an initial speed 10\,\text{м/с} and acceleration 0{,}5\,\text{м/с}^{2}. How long will it take? 44\,\text{м}?

Answer options

  1. 4\,\text{с}
  2. 2\,\text{с}
  3. 8\,\text{с}
  4. 40\,\text{с}
Show the walkthroughHide the walkthrough
Given
v_0
10\,\text{м/с}
a
0{,}5\,\tfrac{\text{м}}{\text{с}^{2}}
S
44\,\text{м}
Find

t

  1. S = v_0 t + \dfrac{a_x t^{2}}{2}\ \Rightarrow\ a_x t^{2} + 2v_0 t - 2S = 0

    We reduce to the usual quadratic equation for time.

  2. t_{1,2} = \dfrac{-2v_0 \pm \sqrt{4v_0^{2} + 8a_x S}}{2a_x}\dfrac{-20 \pm \sqrt{400 + 176}}{2\cdot 0{,}5} = \dfrac{-20 + 24}{1}t = 4\,\text{с}

    We discard the negative root: time cannot be less than zero.

Answert = 4\,\text{с}

Position in this set: 33 from 50

Problem #129

Speed and uniform motion

A cyclist moves uniformly at a speed 4 m/s for 30 s. Find the path you have taken.

Answer options

  1. 120\,\text{м}
  2. 240\,\text{м}
  3. 60\,\text{м}
  4. 360\,\text{м}
Show the walkthroughHide the walkthrough
Given
v
4\,\text{м/с}
t
30\,\text{с}
Find

S

  1. S=vtS=4 \cdot 30S=120\,\text{м}

    In uniform motion, the path is equal to the product of speed and time.

Answer120\,\text{м}

Position in this set: 34 from 50

Problem #130

Speed and uniform motion

A cyclist moves uniformly at a speed 6 m/s for 40 s. Find the path you have taken.

Answer options

  1. 480\,\text{м}
  2. 240\,\text{м}
  3. 120\,\text{м}
  4. 720\,\text{м}
Show the walkthroughHide the walkthrough
Given
v
6\,\text{м/с}
t
40\,\text{с}
Find

S

  1. S=vtS=6 \cdot 40S=240\,\text{м}

    In uniform motion, the path is equal to the product of speed and time.

Answer240\,\text{м}

Position in this set: 35 from 50

Problem #131

Speed and uniform motion

The tourist passed 300 m to the east, then 100 m to the west. Took all the way 200 s. Find the average ground speed.

Answer options

  1. 4\,\text{м/с}
  2. 2\,\text{м/с}
  3. 1\,\text{м/с}
  4. 6\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
a
300\,\text{м}
b
100\,\text{м}
t
200\,\text{с}
Find

v

  1. S=a+bS=300 + 100S=400\,\text{м}

    We add up the path regardless of the direction of movement.

  2. v=\frac{S}{t}v=\frac{400}{200}v=2\,\text{м/с}

    Average ground speed uses the entire path rather than travel.

Answer2\,\text{м/с}

Position in this set: 36 from 50

Problem #132

Speed and uniform motion

The tourist passed 450 m to the east, then 150 m to the west. Took all the way 240 s. Find the average ground speed.

Answer options

  1. 5\,\text{м/с}
  2. 1{,}25\,\text{м/с}
  3. 2{,}5\,\text{м/с}
  4. 7{,}5\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
a
450\,\text{м}
b
150\,\text{м}
t
240\,\text{с}
Find

v

  1. S=a+bS=450 + 150S=600\,\text{м}

    We add up the path regardless of the direction of movement.

  2. v=\frac{S}{t}v=\frac{600}{240}v=2{,}5\,\text{м/с}

    Average ground speed uses the entire path rather than travel.

Answer2{,}5\,\text{м/с}

Position in this set: 37 from 50

Problem #133

Speed and uniform motion

The tourist passed 240 m at speed 2 m/s, resting 60 s and passed some more 360 m at speed 3 m/s. Find the average speed over the entire time.

Answer options

  1. 4\,\text{м/с}
  2. 1\,\text{м/с}
  3. 2\,\text{м/с}
  4. 6\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
a
240\,\text{м}
b
360\,\text{м}
v_1
2\,\text{м/с}
v_2
3\,\text{м/с}
t_0
60\,\text{с}
Find

v

  1. t_1=\frac{a}{v_1}t_1=\frac{240}{2}t_1=120\,\text{с}

    We find the time of the first section.

  2. t_2=\frac{b}{v_2}t_2=\frac{360}{3}t_2=120\,\text{с}

    Finding the time for the second section.

  3. T=t_1+t_0+t_2T=120 + 60 + 120T=300\,\text{с}

    The stop is included in the total time.

  4. v=\frac{a+b}{T}v=\frac{(240 + 360)}{300}v=2\,\text{м/с}

    We divide the total path by the total time.

Answer2\,\text{м/с}

Position in this set: 38 from 50

Problem #134

Speed and uniform motion

The tourist passed 360 m at speed 3 m/s, resting 60 s and passed some more 540 m at speed 4{,}5 m/s. Find the average speed over the entire time.

Answer options

  1. 6\,\text{м/с}
  2. 1{,}5\,\text{м/с}
  3. 9\,\text{м/с}
  4. 3\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
a
360\,\text{м}
b
540\,\text{м}
v_1
3\,\text{м/с}
v_2
4{,}5\,\text{м/с}
t_0
60\,\text{с}
Find

v

  1. t_1=\frac{a}{v_1}t_1=\frac{360}{3}t_1=120\,\text{с}

    We find the time of the first section.

  2. t_2=\frac{b}{v_2}t_2=\frac{540}{4{,}5}t_2=120\,\text{с}

    Finding the time for the second section.

  3. T=t_1+t_0+t_2T=120 + 60 + 120T=300\,\text{с}

    The stop is included in the total time.

  4. v=\frac{a+b}{T}v=\frac{(360 + 540)}{300}v=3\,\text{м/с}

    We divide the total path by the total time.

Answer3\,\text{м/с}

Position in this set: 39 from 50

Problem #135

Uniformly accelerated motion

The body moves with initial speed 2 m/s and constant acceleration 3 m/s² in the same direction. Find the speed through 4 s.

Answer options

  1. 28\,\text{м/с}
  2. 14\,\text{м/с}
  3. 7\,\text{м/с}
  4. 42\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
v_0
2\,\text{м/с}
a
3\,\text{м/с}^{2}
t
4\,\text{с}
Find

v

  1. v=v_0+atv=2 + 3 \cdot 4v=14\,\text{м/с}

    Acceleration and initial velocity are co-directional.

Answer14\,\text{м/с}

Position in this set: 40 from 50

Problem #136

Uniformly accelerated motion

The body moves with initial speed 4 m/s and constant acceleration 2 m/s² in the same direction. Find the speed through 5 s.

Answer options

  1. 28\,\text{м/с}
  2. 7\,\text{м/с}
  3. 14\,\text{м/с}
  4. 42\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
v_0
4\,\text{м/с}
a
2\,\text{м/с}^{2}
t
5\,\text{с}
Find

v

  1. v=v_0+atv=4 + 2 \cdot 5v=14\,\text{м/с}

    Acceleration and initial velocity are co-directional.

Answer14\,\text{м/с}

Position in this set: 41 from 50

Problem #137

Uniformly accelerated motion

Vehicle speed for 8 with increased evenly with 4 up to 20 m/s. Find a way.

Answer options

  1. 192\,\text{м}
  2. 48\,\text{м}
  3. 96\,\text{м}
  4. 288\,\text{м}
Show the walkthroughHide the walkthrough
Given
t
8\,\text{с}
v_0
4\,\text{м/с}
v
20\,\text{м/с}
Find

S

  1. v_{\text{ср}}=\frac{v_0+v}{2}v_{\text{ср}}=\frac{(4 + 20)}{2}v_{\text{ср}}=12\,\text{м/с}

    With constant acceleration, the average speed is equal to half the sum of the extreme ones.

  2. S=v_{\text{ср}}tS=12 \cdot 8S=96\,\text{м}

    Multiply the average speed by the time.

Answer96\,\text{м}

Position in this set: 42 from 50

Problem #138

Uniformly accelerated motion

Vehicle speed for 10 with increased evenly with 6 up to 24 m/s. Find a way.

Answer options

  1. 300\,\text{м}
  2. 75\,\text{м}
  3. 450\,\text{м}
  4. 150\,\text{м}
Show the walkthroughHide the walkthrough
Given
t
10\,\text{с}
v_0
6\,\text{м/с}
v
24\,\text{м/с}
Find

S

  1. v_{\text{ср}}=\frac{v_0+v}{2}v_{\text{ср}}=\frac{(6 + 24)}{2}v_{\text{ср}}=15\,\text{м/с}

    With constant acceleration, the average speed is equal to half the sum of the extreme ones.

  2. S=v_{\text{ср}}tS=15 \cdot 10S=150\,\text{м}

    Multiply the average speed by the time.

Answer150\,\text{м}

Position in this set: 43 from 50

Problem #139

Uniformly accelerated motion

The car is traveling at speed 20 m/s. The driver reacts 0{,}5 s, then the car slows down with acceleration constant in modulus 5 m/s² to the stop. Find the complete stopping path.

Answer options

  1. 100\,\text{м}
  2. 25\,\text{м}
  3. 150\,\text{м}
  4. 50\,\text{м}
Show the walkthroughHide the walkthrough
Given
v
20\,\text{м/с}
t
0{,}5\,\text{с}
a
5\,\text{м/с}^{2}
Find

S

  1. S_1=vtS_1=20 \cdot 0{,}5S_1=10\,\text{м}

    During the reaction, the speed is still constant.

  2. S_2=\frac{v^2}{2a}S_2=\frac{{20}^{2}}{(2 \cdot 5)}S_2=40\,\text{м}

    The braking distance is derived from the dependence of speed on movement.

  3. S=S_1+S_2S=10 + 40S=50\,\text{м}

    The complete path includes both stages.

Answer50\,\text{м}

Position in this set: 44 from 50

Problem #140

Uniformly accelerated motion

The car is traveling at speed 30 m/s. The driver reacts 0{,}6 s, then the car slows down with acceleration constant in modulus 6 m/s² to the stop. Find the complete stopping path.

Answer options

  1. 93\,\text{м}
  2. 186\,\text{м}
  3. 46{,}5\,\text{м}
  4. 279\,\text{м}
Show the walkthroughHide the walkthrough
Given
v
30\,\text{м/с}
t
0{,}6\,\text{с}
a
6\,\text{м/с}^{2}
Find

S

  1. S_1=vtS_1=30 \cdot 0{,}6S_1=18\,\text{м}

    During the reaction, the speed is still constant.

  2. S_2=\frac{v^2}{2a}S_2=\frac{{30}^{2}}{(2 \cdot 6)}S_2=75\,\text{м}

    The braking distance is derived from the dependence of speed on movement.

  3. S=S_1+S_2S=18 + 75S=93\,\text{м}

    The complete path includes both stages.

Answer93\,\text{м}

Position in this set: 45 from 50

Problem #141

Free fall and circular motion

The body is released without initial velocity and falls 3 s. Find the speed before the impact; g = 10 m/s², no air resistance.

Answer options

  1. 60\,\text{м/с}
  2. 15\,\text{м/с}
  3. 30\,\text{м/с}
  4. 90\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
t
3\,\text{с}
g
10\,\text{м/с}^{2}
Find

v

  1. v=gtv=10 \cdot 3v=30\,\text{м/с}

    The initial speed is zero.

Answer30\,\text{м/с}

Position in this set: 46 from 50

Problem #142

Free fall and circular motion

The body is released without initial velocity and falls 4 s. Find the speed before the impact; g = 10 m/s², no air resistance.

Answer options

  1. 80\,\text{м/с}
  2. 20\,\text{м/с}
  3. 120\,\text{м/с}
  4. 40\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
t
4\,\text{с}
g
10\,\text{м/с}^{2}
Find

v

  1. v=gtv=10 \cdot 4v=40\,\text{м/с}

    The initial speed is zero.

Answer40\,\text{м/с}

Position in this set: 47 from 50

Problem #143

Free fall and circular motion

A body falls without initial speed from a height 45 m. Find the time of fall at g = 10 m/s². Neglect air resistance.

Answer options

  1. 6\,\text{с}
  2. 1{,}5\,\text{с}
  3. 9\,\text{с}
  4. 3\,\text{с}
Show the walkthroughHide the walkthrough
Given
h
45\,\text{м}
g
10\,\text{м/с}^{2}
Find

t

  1. t^2=\frac{2h}{g}t^2=\frac{2 \cdot 45}{10}t^2=9\,\text{с}^{2}

    We express the square of time from the law of path.

  2. t=\sqrt{\frac{2h}{g}}t=\sqrt{9}t=3\,\text{с}

    We choose positive time.

Answer3\,\text{с}

Position in this set: 48 from 50

Problem #144

Free fall and circular motion

A body falls without initial speed from a height 80 m. Find the time of fall at g = 10 m/s². Neglect air resistance.

Answer options

  1. 4\,\text{с}
  2. 8\,\text{с}
  3. 2\,\text{с}
  4. 12\,\text{с}
Show the walkthroughHide the walkthrough
Given
h
80\,\text{м}
g
10\,\text{м/с}^{2}
Find

t

  1. t^2=\frac{2h}{g}t^2=\frac{2 \cdot 80}{10}t^2=16\,\text{с}^{2}

    We express the square of time from the law of path.

  2. t=\sqrt{\frac{2h}{g}}t=\sqrt{16}t=4\,\text{с}

    We choose positive time.

Answer4\,\text{с}

Position in this set: 49 from 50

Problem #145

Free fall and circular motion

A body is thrown horizontally with a speed 15 m/s from height 20 m. Find the magnitude of the velocity when falling to the ground. g = 10 m/s², no air resistance.

Answer options

  1. 25\,\text{м/с}
  2. 50\,\text{м/с}
  3. 12{,}5\,\text{м/с}
  4. 75\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
v
15\,\text{м/с}
h
20\,\text{м}
g
10\,\text{м/с}^{2}
Find

V

  1. t=\sqrt{\frac{2h}{g}}t=\sqrt{\frac{2 \cdot 20}{10}}t=2\,\text{с}

    Vertical movement determines flight time.

  2. v_y=gtv_y=10 \cdot 2v_y=20\,\text{м/с}

    The vertical speed increases, the horizontal speed remains constant.

  3. V=\sqrt{v_x^2+v_y^2}V=\sqrt{{15}^{2} + {20}^{2}}V=25\,\text{м/с}

    Add the perpendicular components using the Pythagorean theorem.

Answer25\,\text{м/с}

Position in this set: 50 from 50

Problem #146

Free fall and circular motion

A body is thrown horizontally with a speed 24 m/s from height 12{,}8 m. Find the magnitude of the velocity when falling to the ground. g = 10 m/s², no air resistance.

Answer options

  1. 57{,}6888\,\text{м/с}
  2. 28{,}8444\,\text{м/с}
  3. 14{,}4222\,\text{м/с}
  4. 86{,}5332\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
v
24\,\text{м/с}
h
12{,}8\,\text{м}
g
10\,\text{м/с}^{2}
Find

V

  1. t=\sqrt{\frac{2h}{g}}t=\sqrt{\frac{2 \cdot 12{,}8}{10}}t=1{,}6\,\text{с}

    Vertical movement determines flight time.

  2. v_y=gtv_y=10 \cdot 1{,}6v_y=16\,\text{м/с}

    The vertical speed increases, the horizontal speed remains constant.

  3. V=\sqrt{v_x^2+v_y^2}V=\sqrt{{24}^{2} + {16}^{2}}V=28{,}8444\,\text{м/с}

    Add the perpendicular components using the Pythagorean theorem.

Answer28{,}8444\,\text{м/с}

The theory behind these problems

Work through the topic if a problem will not give in.

Speed and uniform motionDistance, time, speed, average speed and combined motions.Uniformly accelerated motionAcceleration, velocity and displacement under constant acceleration.Free fall and circular motionFall without resistance and the kinematics of uniform rotation.

Formulas by topic

Look up symbols and units of measurement in the reference.

PeriodT · сVelocityv · м/сAngular velocityω · рад/сAccelerationa · м/с²Frequencyν · Гц