Browse Kinematics

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Motion graphs (displacement–time & velocity–time)

20 questions

LessonNot started
A pebble is projected vertically upwards with speed U m s−1U\ \text{m s}^{-1} from a point AA on a bridge. The point AA is 22.422.4 m vertically above the surface of a river.

The pebble moves freely under gravity and hits the surface of the river 44 s after it is projected.
(a) Find the value of UU.

(3 marks)
(b) Find the speed of the pebble at the instant it hits the surface of the river.

(3 marks)
(c) Sketch a speed-time graph for the motion of the pebble from the instant it is projected to the instant it hits the surface of the river.

(No further calculations are required.)

(2 marks)
●●●●●Level 38 marksStart
A train travels between two stations 20002000 m apart. It starts from rest and accelerates uniformly at 0.80.8 m s⁻² for 2525 s, then travels at constant speed for TT seconds, and finally decelerates uniformly at 1.251.25 m s⁻² to rest at the second station.
(a) Sketch a velocity–time graph for the journey, labelling the key values. (2 marks)
(b) Find the value of TT. (4 marks)
(c) Find the average speed of the train for the whole journey. (2 marks)
(d) Explain why the average speed is less than the constant speed reached. (1 mark)
●●●●●Level 39 marksStart
SECTION B: MECHANICS

A train travels between two stations. It starts from rest and accelerates uniformly to a speed of 3030 m s⁻¹ in 2020 seconds. It then travels at 3030 m s⁻¹ for TT seconds, before decelerating uniformly to rest in a further 1515 seconds.

The total distance between the stations is 24002400 m.
(a) Find the total distance travelled during the acceleration and deceleration phases. (2 marks)
(b) Find the value of TT. (2 marks)
(c) Find the average speed of the train for the whole journey, giving your answer to 33 significant figures. (2 marks)
●●●●●Level 36 marksStart
A remote-controlled car moves in a straight line along a horizontal track.

The graph below shows how the car's velocity, v m s−1v\ \text{m s}^{-1}, changes with time, tt seconds.12206−6tvOver the period 0≤t≤200 \le t \le 20 the car has a total displacement of −30-30 metres.

Initially the car has velocity 0 m s−10\ \text{m s}^{-1}

Find the next time when the velocity of the car is 0 m s−10\ \text{m s}^{-1}

(4 marks)
●●●●●Level 34 marksStart
The velocity-time graph shows a particle's motion. Find the total distance travelled.246810125101520t (s)v (m/s)
●●●●●Level 32 marksStart
A velocity-time graph shows a particle accelerating uniformly from rest to 20 m s−120\ \text{m s}^{-1} in 44 s. Find the acceleration.
●●●●●Level 32 marksStart
The velocity-time graph shows a particle's journey. Find
(a) the total distance travelled and (b) the average speed.510152051015t (s)v (m/s)
●●●●●Level 33 marksStart
A particle decelerates uniformly from 18 m s−118\ \text{m s}^{-1} to 6 m s−16\ \text{m s}^{-1} in 33 s. Find the deceleration (the magnitude of the gradient of the velocity-time graph).
●●●●●Level 32 marksStart
On a velocity-time graph a particle travels at a constant 12 m s−112\ \text{m s}^{-1} for 88 s, then decelerates to rest in a further 44 s. Find the total distance travelled.
●●●●●Level 32 marksStart
A particle accelerates from rest at 3 m s−23\ \text{m s}^{-2} for 44 s, then travels at constant velocity for 66 s. Find the total distance travelled.
●●●●●Level 32 marksStart

Constant acceleration (suvat)

54 questions

LessonNot started
A particle PP moves with constant acceleration (1.5i−4j) m s−2(1.5\mathbf{i} - 4\mathbf{j})\ \text{m s}^{-2}. At time t=0t = 0 seconds, PP is moving with velocity (−2i+7j) m s−1(-2\mathbf{i} + 7\mathbf{j})\ \text{m s}^{-1}.
(a) Determine the speed of PP when t=4t = 4.

(4 marks)
(b) Find, in terms of i\mathbf{i} and j\mathbf{j}, the displacement of PP from its position at t=0t = 0 to its position at t=4t = 4.

(2 marks)
●●●●●Level 36 marksStart
A train travels between two stations 20002000 m apart. It starts from rest and accelerates uniformly at 0.80.8 m s⁻² for 2525 s, then travels at constant speed for TT seconds, and finally decelerates uniformly at 1.251.25 m s⁻² to rest at the second station.
(a) Sketch a velocity–time graph for the journey, labelling the key values. (2 marks)
(b) Find the value of TT. (4 marks)
(c) Find the average speed of the train for the whole journey. (2 marks)
(d) Explain why the average speed is less than the constant speed reached. (1 mark)
●●●●●Level 39 marksStart
SECTION B: MECHANICS

A train travels between two stations. It starts from rest and accelerates uniformly to a speed of 3030 m s⁻¹ in 2020 seconds. It then travels at 3030 m s⁻¹ for TT seconds, before decelerating uniformly to rest in a further 1515 seconds.

The total distance between the stations is 24002400 m.
(a) Find the total distance travelled during the acceleration and deceleration phases. (2 marks)
(b) Find the value of TT. (2 marks)
(c) Find the average speed of the train for the whole journey, giving your answer to 33 significant figures. (2 marks)
●●●●●Level 36 marksStart
A remote-controlled car moves in a straight line along a horizontal track.

The graph below shows how the car's velocity, v m s−1v\ \text{m s}^{-1}, changes with time, tt seconds.12206−6tvOver the period 0≤t≤200 \le t \le 20 the car has a total displacement of −30-30 metres.

Initially the car has velocity 0 m s−10\ \text{m s}^{-1}

Find the next time when the velocity of the car is 0 m s−10\ \text{m s}^{-1}

(4 marks)
●●●●●Level 34 marksStart
A motorcyclist has an initial velocity of 2 m s−12\ \text{m s}^{-1}

The motorcyclist is moving in a straight line along a horizontal road.

The acceleration, a m s−2a\ \text{m s}^{-2}, of the motorcyclist at time tt seconds is given by

a=1.2−4kt+9kt2a = 1.2 - 4kt + 9kt^2

where kk is a constant.

When t=5t = 5 the motorcyclist has a velocity of 21 m s−121\ \text{m s}^{-1}

Show that k=125k = \dfrac{1}{25}

(4 marks)
●●●●●Level 34 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

Two drops of rainwater hang, at rest, from the edges of two different balconies on the same side of a block of flats. The ground below the balconies is horizontal.

The vertical distance between the two drops is dd metres.

At the same instant both drops begin to fall freely under gravity.

The first drop hits the ground after 1.61.6 seconds.

The second drop hits the ground 0.30.3 seconds later.

Show that dd is approximately 5.15.1

(4 marks)
●●●●●Level 34 marksStart
A particle accelerates uniformly from 5 m s−15\ \text{m s}^{-1} to 15 m s−115\ \text{m s}^{-1} with acceleration 2 m s−22\ \text{m s}^{-2}. Find the distance travelled.
●●●●●Level 32 marksStart
A particle has initial velocity 10 m s−110\ \text{m s}^{-1} and acceleration −2 m s−2-2\ \text{m s}^{-2}. Find its displacement after 33 s.
●●●●●Level 32 marksStart
A car accelerates uniformly from 4 m s−14\ \text{m s}^{-1} to 20 m s−120\ \text{m s}^{-1} in 88 s. Find its acceleration.
●●●●●Level 32 marksStart
A particle starts at 8 m s−18\ \text{m s}^{-1} and travels 100100 m in 1010 s under constant acceleration. Find the acceleration.
●●●●●Level 32 marksStart
A train travelling at 30 m s−130\ \text{m s}^{-1} brakes uniformly and stops in 600600 m. Find the deceleration.
●●●●●Level 32 marksStart

Vertical motion under gravity

14 questions

LessonNot started
A parcel of mass 1.51.5 kg is released from rest from a drone that is hovering at a height of 4040 m above horizontal ground.

In a first model, the parcel is modelled as a particle moving freely under gravity.
(a) Using this first model, find the time taken for the parcel to reach the ground.

(3 marks)

In a second model, the parcel is modelled as a particle falling against a constant resistance to motion of magnitude RR newtons. Using this second model, the parcel takes 3.53.5 s to reach the ground.
(b) Find the value of RR.

(5 marks)
●●●●●Level 38 marksStart
A pebble is projected vertically upwards with speed U m s−1U\ \text{m s}^{-1} from a point AA on a bridge. The point AA is 22.422.4 m vertically above the surface of a river.

The pebble moves freely under gravity and hits the surface of the river 44 s after it is projected.
(a) Find the value of UU.

(3 marks)
(b) Find the speed of the pebble at the instant it hits the surface of the river.

(3 marks)
(c) Sketch a speed-time graph for the motion of the pebble from the instant it is projected to the instant it hits the surface of the river.

(No further calculations are required.)

(2 marks)
●●●●●Level 38 marksStart
A small ball of mass 0.20.2 kg is released from rest at a point 2.52.5 m above a horizontal floor. The ball falls freely under gravity, hits the floor and rebounds vertically upwards. The coefficient of restitution between the ball and the floor is 0.60.6

The ball is modelled as a particle.
(a) Find the speed of the ball immediately before it hits the floor.

(2 marks)
(b) Find the magnitude of the impulse exerted on the ball by the floor.

(3 marks)
(c) Find the greatest height above the floor reached by the ball after it rebounds.

(2 marks)
●●●●●Level 37 marksStart
In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

Two drops of rainwater hang, at rest, from the edges of two different balconies on the same side of a block of flats. The ground below the balconies is horizontal.

The vertical distance between the two drops is dd metres.

At the same instant both drops begin to fall freely under gravity.

The first drop hits the ground after 1.61.6 seconds.

The second drop hits the ground 0.30.3 seconds later.

Show that dd is approximately 5.15.1

(4 marks)
●●●●●Level 34 marksStart
A ball is thrown vertically upwards at 14.7 m s−114.7\ \text{m s}^{-1}. Find the time to reach its highest point. (g=9.8g = 9.8.)
●●●●●Level 32 marksStart
A ball is thrown vertically upwards at 19.6 m s−119.6\ \text{m s}^{-1}. Find the greatest height reached. (g=9.8g = 9.8.)
●●●●●Level 32 marksStart
A ball is thrown vertically upwards at 24.5 m s−124.5\ \text{m s}^{-1} from ground level. Find the total time before it returns to the ground. (g=9.8g = 9.8.)
●●●●●Level 32 marksStart
A stone is dropped from a height of 44.144.1 m. Find the time it takes to reach the ground. (g=9.8g = 9.8.)
●●●●●Level 32 marksStart
A ball is thrown vertically downwards at 5 m s−15\ \text{m s}^{-1} from a window. Find
(a) its speed and (b) the distance fallen after 22 s. (g=9.8g = 9.8.)
●●●●●Level 34 marksStart

Kinematics with calculus (a → v → s)

34 questions

LessonNot started
A particle PP moves in a straight line. At time tt seconds (0≤t≤40 \le t \le 4), its velocity vv m s⁻¹ is given by

v=12t−3t2v = 12t - 3t^2.
(a) Find the acceleration of PP when t=3t = 3. (2 marks)
(b) Find the displacement of PP from its starting position when t=4t = 4. (3 marks)
(c) Determine the maximum velocity of PP. (2 marks)
●●●●●Level 37 marksStart
A motorcyclist has an initial velocity of 2 m s−12\ \text{m s}^{-1}

The motorcyclist is moving in a straight line along a horizontal road.

The acceleration, a m s−2a\ \text{m s}^{-2}, of the motorcyclist at time tt seconds is given by

a=1.2−4kt+9kt2a = 1.2 - 4kt + 9kt^2

where kk is a constant.

When t=5t = 5 the motorcyclist has a velocity of 21 m s−121\ \text{m s}^{-1}

Show that k=125k = \dfrac{1}{25}

(4 marks)
●●●●●Level 34 marksStart
A particle moves in a straight line with acceleration a=(6t−4) m s−2a = (6t - 4)\ \text{m s}^{-2}. At t=0t = 0 its velocity is 2 m s−12\ \text{m s}^{-1}.
(a) Find an expression for the velocity at time tt. (3 marks)
(b) Find the velocity when t=2t = 2. (1 mark)
●●●●●Level 34 marksStart
A particle moves in a straight line with velocity v=(2t+3) m s−1v = (2t + 3)\ \text{m s}^{-1}. At t=0t = 0 its displacement from a fixed point OO is 5 m5\text{ m}.
(a) Find an expression for the displacement from OO at time tt. (3 marks)
(b) Find the displacement when t=4t = 4. (1 mark)
●●●●●Level 34 marksStart
A particle moves in a straight line with velocity v=(t2−5t+6) m s−1v = (t^{2} - 5t + 6)\ \text{m s}^{-1} at time tt seconds.
(a) Find the times at which the particle is instantaneously at rest. (2 marks)
(b) Find the displacement of the particle from its starting point when t=4t = 4. (3 marks)
(c) Find the total distance travelled between t=0t = 0 and t=4t = 4. (4 marks)
●●●●●Level 39 marksStart
A particle moves in a straight line with velocity v=12(t+1)2 m s−1v = \dfrac{12}{(t + 1)^{2}}\ \text{m s}^{-1} at time tt seconds.
(a) Find the acceleration when t=1t = 1. (3 marks)
(b) Find the distance travelled between t=0t = 0 and t=1t = 1. (3 marks)
●●●●●Level 36 marksStart
A particle moves in a straight line with velocity v=(12t−3t2) m s−1v = (12t - 3t^{2})\ \text{m s}^{-1} at time tt seconds.

Find the maximum velocity of the particle. (4 marks)
●●●●●Level 34 marksStart
A particle sets off from rest and moves in a straight line. For 0⩽t⩽20 \leqslant t \leqslant 2 its velocity is v=3t2 m s−1v = 3t^{2}\ \text{m s}^{-1}; for t>2t > 2 it moves at the constant velocity it has reached at t=2t = 2.

Find the total distance travelled in the first 55 seconds. (5 marks)
●●●●●Level 35 marksStart
A particle starts from rest at a point OO and moves in a straight line with acceleration a=(6−2t) m s−2a = (6 - 2t)\ \text{m s}^{-2} at time tt seconds.
(a) Find an expression for the velocity at time tt, and the maximum velocity. (3 marks)
(b) Find the time at which the particle returns to instantaneous rest. (1 mark)
(c) Find the distance travelled up to that time. (2 marks)
●●●●●Level 36 marksStart

Motion in two dimensions with vectors

48 questions

LessonNot started
[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A particle PP of mass 0.80.8 kg is moving on a smooth horizontal surface with velocity (3i+2j) m s−1(3\mathbf{i} + 2\mathbf{j})\text{ m s}^{-1} when it receives an impulse λ(2i−j)\lambda(2\mathbf{i} - \mathbf{j}) N s, where λ\lambda is a constant. Immediately after receiving the impulse, PP is moving parallel to i\mathbf{i}.
(a) Find the value of λ\lambda.

(4 marks)
(b) Find the speed of PP immediately after receiving the impulse.

(2 marks)
●●●●●Level 36 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal perpendicular unit vectors.]

A puck of mass 0.050.05 kg glides across the smooth horizontal surface of an air-hockey table with velocity (6i+2j) m s−1(6\mathbf{i} + 2\mathbf{j})\text{ m s}^{-1}. A player hits the puck, and it then moves with velocity (−6i+11j) m s−1(-6\mathbf{i} + 11\mathbf{j})\text{ m s}^{-1}. The puck is modelled as a particle.
(a) Find the magnitude of the impulse that the player gives the puck.

(4 marks)
(b) Find the angle between the direction of this impulse and the direction in which the puck was moving before it was hit.

(3 marks)
●●●●●Level 37 marksStart
A particle PP moves with constant acceleration (1.5i−4j) m s−2(1.5\mathbf{i} - 4\mathbf{j})\ \text{m s}^{-2}. At time t=0t = 0 seconds, PP is moving with velocity (−2i+7j) m s−1(-2\mathbf{i} + 7\mathbf{j})\ \text{m s}^{-1}.
(a) Determine the speed of PP when t=4t = 4.

(4 marks)
(b) Find, in terms of i\mathbf{i} and j\mathbf{j}, the displacement of PP from its position at t=0t = 0 to its position at t=4t = 4.

(2 marks)
●●●●●Level 36 marksStart
[In this question i\mathbf{i} and j\mathbf{j} are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin OO.]

Two remote-controlled model yachts, PP and QQ, are sailing on a large lake. Each yacht is moving with constant velocity and the yachts are modelled as particles.

At time tt seconds, the position vector of PP is r\mathbf{r} metres.
  • When t=4t = 4, r=(−3i+14j)\mathbf{r} = (-3\mathbf{i} + 14\mathbf{j})
  • When t=10t = 10, r=(15i+2j)\mathbf{r} = (15\mathbf{i} + 2\mathbf{j})
(a) Find, in terms of i\mathbf{i} and j\mathbf{j}, the velocity of PP.

(2 marks)
(b) Find an expression for r\mathbf{r} in terms of tt, i\mathbf{i} and j\mathbf{j}.

(2 marks)

At time tt seconds, the position vector of QQ is s\mathbf{s} metres, where

s=8i−5j+t(−i+4j)\mathbf{s} = 8\mathbf{i} - 5\mathbf{j} + t(-\mathbf{i} + 4\mathbf{j})
(c) Find, to the nearest degree, the bearing of QQ from PP when t=10t = 10

(3 marks)
●●●●●Level 37 marksStart
A particle moves in a plane so that at time tt seconds its position vector is r=(t3−2t) i+(3t2−t) j\mathbf{r} = (t^3 - 2t)\,\mathbf{i} + (3t^2 - t)\,\mathbf{j} metres. Find
(a) its velocity at time tt,
(b) its acceleration at time tt, and (c) its speed when t=1t = 1.
●●●●●Level 35 marksStart
A particle moves with constant acceleration a=(−i+2j) m s−2\mathbf{a} = (-\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-2}. At t=0t = 0 its velocity is u=(4i−3j) m s−1\mathbf{u} = (4\mathbf{i} - 3\mathbf{j})\ \text{m s}^{-1}. Find
(a) its velocity when t=3t = 3,
(b) its speed when t=3t = 3, and (c) the value of tt at which it is moving parallel to j\mathbf{j}.
●●●●●Level 34 marksStart
A particle moves in a plane so that at time tt seconds its position vector is

r=(3t2−1) i+(2t+4) j metres\mathbf{r} = (3t^{2} - 1)\,\mathbf{i} + (2t + 4)\,\mathbf{j} \text{ metres}

Find the speed of the particle when t=2t = 2, giving your answer in the form k37k\sqrt{37} m s−1^{-1}. (4 marks)
●●●●●Level 34 marksStart
A particle has initial velocity (2i−3j)(2\mathbf{i} - 3\mathbf{j}) m s−1^{-1} and moves with constant acceleration (4i+j)(4\mathbf{i} + \mathbf{j}) m s−2^{-2}.

Find its velocity after 33 seconds, and describe the direction of motion at that instant. (3 marks)
●●●●●Level 33 marksStart