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GRADE 9 PRACTICE

Physics Problems: Kinematics, Grade 9

Coordinates, vectors, graphs, acceleration and circular motion.

27 problems across 5 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.

27 problems
Position in this set: 1 from 27

Problem #1

Frame of reference and uniform motion

What concept does the description correspond to: “A model of a body whose size and shape can be neglected when solving a chosen problem.”?

Answer options

  1. \text{Point particle}
  2. \text{Displacement and its component}
  3. \text{Position equation}
  4. \text{Instantaneous velocity}
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Given
\text{Description}
\text{A model of an object whose size and shape can be neglected for the problem being solved.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    A car can be treated as a point on a long highway, but its dimensions matter when parking in a narrow garage.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Point particle}

Answer\text{Concept} = \text{Point particle}

Position in this set: 2 from 27

Problem #2

Frame of reference and uniform motion

What concept does the description correspond to: “Displacement is a vector from the initial position to the final position, and the projection specifies its component on the axis.”?

Answer options

  1. \text{Displacement and its component}
  2. \text{Point particle}
  3. \text{Position equation}
  4. \text{Instantaneous velocity}
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Given
\text{Description}
\text{Displacement is a vector from the initial position to the final position. Its component gives the signed value along a chosen axis.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    A component can be positive, negative or zero. Distance travelled remains the nonnegative length of the path.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Displacement and its component}

Answer\text{Concept} = \text{Displacement and its component}

Position in this set: 3 from 27

Problem #3

Kinematics

The body begins to move from point x₀ = -10 m with a constant velocity projection of 4 m/s. Find the coordinate after 5 s.

Answer options

  1. 10\,\text{м}
  2. -30\,\text{м}
  3. -6\,\text{м}
  4. 30\,\text{м}
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Given
x_0
-10\,\text{м}
v_x
4\,\text{м/с}
t
5\,\text{с}
Find

x

  1. x=x_0+v_xtx=-10+4\cdot5\,\text{м}x = 10\,\text{м}

Answer10\,\text{м}

Position in this set: 4 from 27

Problem #4

Uniformly accelerated motion and its graphs

What concept does the description correspond to: “The speed of a body at a given moment in time, directed tangentially to the trajectory.”?

Answer options

  1. \text{Instantaneous velocity}
  2. \text{Graphical methods in kinematics}
  3. \text{Uniformly accelerated motion}
  4. \text{Point particle}
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Given
\text{Description}
\text{An object's velocity at a given instant, directed along the tangent to its path.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Average speed is total distance divided by total time. Average velocity uses the displacement vector, while a speedometer shows instantaneous speed.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Instantaneous velocity}

Answer\text{Concept} = \text{Instantaneous velocity}

Position in this set: 5 from 27

Problem #5

Uniformly accelerated motion and its graphs

What concept does the description correspond to: “The slope of the coordinate graph determines the projection of velocity. The algebraic area under the velocity projection graph is equal to the displacement projection.”?

Answer options

  1. \text{Graphical methods in kinematics}
  2. \text{Instantaneous velocity}
  3. \text{Uniformly accelerated motion}
  4. \text{Point particle}
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Given
\text{Description}
\text{The slope of a coordinate–time graph gives the velocity component. The signed area under a velocity-component–time graph gives the displacement component.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    The slope of a velocity-component graph gives the acceleration component. Speed increases when velocity and acceleration have the same sign, and decreases when their signs are opposite. The slope's sign alone does not tell us whether the object is speeding up.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Graphical methods in kinematics}

Answer\text{Concept} = \text{Graphical methods in kinematics}

Position in this set: 6 from 27

Problem #6

Kinematics

The initial speed of the body is 2 m/s, acceleration is 3 m/s², acceleration time is 4 s. Find the final velocity and displacement.

Answer options

  1. 14\,\text{м/с};\ 32\,\text{м}
  2. 12\,\text{м/с};\ 24\,\text{м}
  3. 14\,\text{м/с};\ 56\,\text{м}
  4. 5\,\text{м/с};\ 20\,\text{м}
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Given
v_0
2\,\text{м/с}
a
3\,\text{м/с}^{2}
t
4\,\text{с}
Find

vs

  1. v=v_0+atv=2+3\cdot4\,\text{м/с}v = 14\,\text{м/с}
  2. s=v_0t+\frac{at^2}{2}s=2\cdot4+\frac{3\cdot4^2}{2}\,\text{м}s = 32\,\text{м}

Answer14\,\text{м/с}32\,\text{м}

Position in this set: 7 from 27

Problem #7

Free fall, circular motion and relative velocity

What concept does the description correspond to: “Motion with a constant velocity modulus, but a continuously changing direction of the velocity vector.”?

Answer options

  1. \text{Uniform circular motion}
  2. \text{Free fall}
  3. \text{Relative velocity}
  4. \text{Point particle}
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Given
\text{Description}
\text{Motion at constant speed with a continuously changing direction of the velocity vector.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    Centripetal acceleration points towards the centre and changes the direction of the velocity vector.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Uniform circular motion}

Answer\text{Concept} = \text{Uniform circular motion}

Position in this set: 8 from 27

Problem #8

Free fall, circular motion and relative velocity

What concept does the description correspond to: “The movement of a body only under the influence of gravity with negligible resistance of the medium.”?

Answer options

  1. \text{Free fall}
  2. \text{Uniform circular motion}
  3. \text{Relative velocity}
  4. \text{Point particle}
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Given
\text{Description}
\text{Motion under gravity alone, with negligible resistance from the surrounding medium.}
Find

\text{name concepts}

  1. \text{Let's compare approval with definition}

    At the same location, all freely falling objects have the same g. It can be measured from falling time and distance or from a pendulum's period, with measurement uncertainty taken into account.

  2. \text{Let's formulate conclusion}\text{Concept} = \text{Free fall}

Answer\text{Concept} = \text{Free fall}

Position in this set: 9 from 27

Problem #9

Circular movement

A car passes a circular turn with a radius of 20 m at a speed of 10 m/s. Find the centripetal acceleration.

Answer options

  1. 5\,\text{м/с}^{2}
  2. 0{,}5\,\text{м/с}^{2}
  3. 2\,\text{м/с}^{2}
  4. 20\,\text{м/с}^{2}
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Given
v
10\,\text{м/с}
r
20\,\text{м}
Find

a_n

  1. a_n=\frac{v^2}{r}a_n=\frac{10^2}{20}\,\text{м/с}^{2}a_n = 5\,\text{м/с}^{2}

Answer5\,\text{м/с}^{2}

Position in this set: 10 from 27

Problem #64

Frame of reference 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}
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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: 11 from 27

Problem #65

Frame of reference 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}
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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: 12 from 27

Problem #66

Frame of reference 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}
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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: 13 from 27

Problem #67

Frame of reference 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}
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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: 14 from 27

Problem #68

Frame of reference 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}
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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: 15 from 27

Problem #69

Frame of reference 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}
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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: 16 from 27

Problem #70

Uniformly accelerated motion and its graphs

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}
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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: 17 from 27

Problem #71

Uniformly accelerated motion and its graphs

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}
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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: 18 from 27

Problem #72

Uniformly accelerated motion and its graphs

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}
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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: 19 from 27

Problem #73

Uniformly accelerated motion and its graphs

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}
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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: 20 from 27

Problem #74

Uniformly accelerated motion and its graphs

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}
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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: 21 from 27

Problem #75

Uniformly accelerated motion and its graphs

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}
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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: 22 from 27

Problem #76

Free fall, circular motion and relative velocity

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}
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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: 23 from 27

Problem #77

Free fall, circular motion and relative velocity

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}
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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: 24 from 27

Problem #78

Free fall, circular motion and relative velocity

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}
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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: 25 from 27

Problem #79

Free fall, circular motion and relative velocity

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}
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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: 26 from 27

Problem #80

Free fall, circular motion and relative velocity

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}
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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: 27 from 27

Problem #81

Free fall, circular motion and relative velocity

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}
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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.

Frame of reference and uniform motionPoint particle, coordinate, displacement, vector components and graphs.Uniformly accelerated motion and its graphsAcceleration, instantaneous velocity, displacement and graphical methods.Free fall, circular motion and relative velocityFalling, centripetal acceleration and velocity relative to different observers.

Formulas by topic

Look up symbols and units of measurement in the reference.

Timet, τ · сDistance and pathS, l · мVelocityv · м/сAccelerationa · м/с²