Questions on kinematics with constant acceleration

Question 1

List the five suvat quantities, and name the equation that leaves out s.

Accepted Answer

Displacement s, initial velocity u, final velocity v, acceleration a and time t. The equation without s is v = u + at. B1 for the list, B1 for the equation. Each of the five standard equations omits exactly one quantity, so choose by the letter that never appears in the question.

Question 2

A car accelerates uniformly from rest at 2 m s⁻² for 6 s. Find its final speed and the distance it covers.

Accepted Answer

Taking u = 0, v = u + at = 0 + 2 × 6 = 12 m s⁻¹, and s = ut + ½at² = 0 + ½ × 2 × 36 = 36 m. M1 A1 for the speed, M1 A1 for the distance. As a check, the average speed over the six seconds is (0 + 12)/2 = 6 m s⁻¹ and 6 × 6 = 36 m, which agrees.

Question 3

A particle moves with constant acceleration a = (i + 4j) m s⁻². At t = 0 its velocity is (3i − 2j) m s⁻¹. Find its velocity when t = 2, and its speed at that instant.

Accepted Answer

v = u + at holds as a vector equation, so v = (3i − 2j) + 2(i + 4j) = (5i + 6j) m s⁻¹. Read component by component it is two scalar equations: i gives 3 + 1 × 2 = 5, j gives −2 + 4 × 2 = 6. The speed is √(5² + 6²) = √61 = 7.81 m s⁻¹ (3 s.f.). M1 for v = u + at applied to each component, A1 for 5i + 6j, A1 for 7.81.

Question 4

The velocity of a particle moving in a straight line increases uniformly from 4 m s⁻¹ to 14 m s⁻¹ in 10 s. Using a velocity-time graph or otherwise, find the acceleration and the displacement of the particle in those 10 s.

Accepted Answer

The graph is a straight line from (0, 4) to (10, 14). Its gradient is (14 − 4)/10 = 1 m s⁻², and the displacement is the area under it, a trapezium: ½(4 + 14) × 10 = 90 m. B1 for the acceleration, B1 for the displacement. Multiplying 14 by 10 uses the final velocity for the whole interval and gives 140 m; the area is a trapezium, not a rectangle.

Question 5

A car brakes uniformly from 30 m s⁻¹ to rest over a distance of 150 m. Find the deceleration and the time taken.

Accepted Answer

Time is not mentioned in the first part, so use v² = u² + 2as: 0 = 900 + 300a, giving a = −3 m s⁻², a deceleration of 3 m s⁻². Then v = u + at gives 0 = 30 − 3t, so t = 10 s. M1 A1 for the acceleration, M1 A1 for the time. Quoting a deceleration of −3 m s⁻² says the car is speeding up.

Question 6

A stone is dropped from rest down a well and hits the water 2.5 s later. Taking g = 9.8 m s⁻², find the depth of the well and the speed of the stone as it hits the water.

Accepted Answer

Taking downwards as positive, u = 0 and a = 9.8. Depth: s = ½gt² = ½ × 9.8 × 6.25 = 30.625, so 30.6 m to 3 significant figures. Impact speed: v = gt = 9.8 × 2.5 = 24.5 m s⁻¹. M1 A1 for the depth, M1 A1 for the speed. 'Dropped' supplies u = 0 without saying so.

Question 7

A ball is thrown vertically upwards from ground level with speed 14 m s⁻¹. Taking g = 9.8 m s⁻², find the greatest height it reaches and the total time it spends in the air before returning to the ground.

Accepted Answer

Take up as positive, so a = −9.8 throughout. At the highest point v = 0, and v² = u² + 2as gives 0 = 196 − 19.6s, so s = 10 m. Time to the top is 14/9.8 = 1.43 s, and by symmetry the whole flight takes twice that, 2.86 s. M1 A1 for the height, M1 A1 for the time. Taking g as positive while taking upwards as positive returns a negative height.

Question 8

A ball is thrown vertically upwards from ground level with speed 14 m s⁻¹. Taking g = 9.8 m s⁻², find the times at which the ball is 5 m above the ground, and explain why there are two of them.

Accepted Answer

With up positive, 5 = 14t − 4.9t², which rearranges to 4.9t² − 14t + 5 = 0. The discriminant is 196 − 98 = 98, so t = (14 ± √98)/9.8, giving t = 0.418 s and t = 2.44 s. M1 for forming the quadratic, M1 for solving it, A1 A1 for the two roots, B1 for the explanation. Two roots mean two visits, since the ball passes 5 m on the way up and again on the way down, so the second root is a physically real answer and is not to be discarded.

Question 9

A particle P moves with constant acceleration a = (0.4i − 1.2j) m s⁻². It passes through the origin O with velocity (2i + 5j) m s⁻¹. Find the position vector of P five seconds later, and its distance from O at that time.

Accepted Answer

r = ut + ½at² as a vector equation gives two component equations: x = 2t + 0.2t² and y = 5t − 0.6t². At t = 5, x = 10 + 5 = 15 and y = 25 − 15 = 10, so r = (15i + 10j) m. The distance from O is √(15² + 10²) = √325 = 18.0 m (3 s.f.). M1 for r = ut + ½at² applied to each component, A1 for both component expressions, A1 for 15i + 10j, A1 for 18.0 m. The j component of the velocity has changed sign by t = 5, since 5 − 1.2 × 5 = −1, but the j displacement is still positive; position and velocity are different vectors and the question asked for position.

Question 10

A train starts from rest and accelerates uniformly to 20 m s⁻¹ in 40 s. It then travels at 20 m s⁻¹ for T seconds, and finally decelerates uniformly to rest in 20 s. The total distance travelled is 2000 m. Sketch the velocity-time graph and find T.

Accepted Answer

The graph is a trapezium: a line from (0, 0) to (40, 20), a horizontal section of length T at 20, then a line down to zero 20 s later. B1 for the sketch with the three sections labelled. The two triangles have areas ½ × 40 × 20 = 400 m and ½ × 20 × 20 = 200 m, and the rectangle has area 20T. So 600 + 20T = 2000, giving 20T = 1400 and T = 70 s. M1 for the areas of the two triangles, A1 for 600 m, M1 for equating the total area to 2000, A1 for T = 70.

Question 11

A particle moves in a straight line with constant acceleration. It passes a point A with speed u m s⁻¹ and, 3 s later, passes a point B with speed 12 m s⁻¹, where AB = 27 m. Find u and the acceleration.

Accepted Answer

The equation that uses s, u, v and t without a is s = ½(u + v)t: 27 = ½(u + 12) × 3, so u + 12 = 18 and u = 6 m s⁻¹. Then v = u + at gives 12 = 6 + 3a, so a = 2 m s⁻². M1 for s = ½(u + v)t, A1 for u = 6, M1 for v = u + at with that u, A1 for a = 2. Choosing s = ½(u + v)t first keeps the unknowns apart: s = ut + ½at² would give one equation in both u and a and need v = u + at solved alongside it, reaching the same answer by a longer route.

Question 12

The graph shows the velocity of a train travelling along a straight track during a 50 s interval. Use the graph to find the acceleration of the train during the first stage of the motion, the total distance the train travels in the 50 s, and its average speed over the whole interval.

Accepted Answer

The first stage runs from the origin to the corner at t = 10 s, v = 20 m s⁻¹, and both of those are read off the gridlines. The acceleration is the gradient of that segment: (20 − 0)/(10 − 0) = 2 m s⁻². M1 for the gradient from two points read off the line, A1 for 2 m s⁻².
Distance is the area between the line and the time axis, and the line stays above the axis throughout, so the whole area counts positively. Split it at the corners into four pieces: a triangle from 0 to 10 s, ½ × 10 × 20 = 100 m; a rectangle from 10 to 30 s, 20 × 20 = 400 m; a trapezium from 30 to 40 s, ½(20 + 10) × 10 = 150 m; and a rectangle from 40 to 50 s, 10 × 10 = 100 m. M1 for splitting the area at the corners and using the right shape for each piece. The total is 100 + 400 + 150 + 100 = 750 m. A1 for 750 m.
Average speed is the total distance divided by the total time, 750/50 = 15 m s⁻¹. A1 for the average speed. It is not the mean of the corner velocities, and it is not the velocity at the halfway time either; only the area settles it. The third segment has gradient (10 − 20)/(40 − 30) = −1 m s⁻², a deceleration half the size of the opening acceleration, which is the sense check that the trapezium, not a second triangle, belongs there.

Question 13

A ball is projected vertically upwards from ground level with speed 14 m s⁻¹. Exactly 1 s later a second ball is projected vertically upwards from the same point with the same speed. Taking g = 9.8 m s⁻², find the time after the first projection at which the two balls are at the same height, and find that height.

Accepted Answer

Measure t from the first projection, with up positive. The first ball is at h₁ = 14t − 4.9t². The second has been in the air for t − 1 seconds, so h₂ = 14(t − 1) − 4.9(t − 1)². Setting them equal, the t² terms cancel and expanding leaves 0 = −14 + 9.8t − 4.9, so 9.8t = 18.9 and t = 1.93 s. Substituting back, h = 14(1.9286) − 4.9(1.9286)² = 8.78 m. M1 for the first height, M1 for using t − 1 in the second, M1 for equating, A1 for the linear equation, A1 for t, A1 for the height. Writing 14t for the second ball as well makes the two expressions identical and the equation unsolvable. Both balls land at t = 2.86 s and t = 3.86 s respectively, so 1.93 s is genuinely during the flight of both.

Question 14

A car travelling along a straight road at a constant 20 m s⁻¹ passes a stationary motorcycle. The motorcycle sets off immediately, accelerating uniformly at 2.5 m s⁻² until it reaches 30 m s⁻¹, after which it maintains that speed. Find the time, from the moment the car passed, at which the motorcycle draws level with the car, and the distance it has travelled by then.

Accepted Answer

Deal with the two phases separately. Phase one: v = u + at gives 30 = 0 + 2.5t, so the motorcycle reaches 30 m s⁻¹ after 12 s, having covered s = ½ × 2.5 × 12² = 180 m. In the same 12 s the car has covered 20 × 12 = 240 m, so it is still 60 m ahead. Phase two: both now travel at constant speed, and the motorcycle closes at 30 − 20 = 10 m s⁻¹, so it takes a further 60/10 = 6 s. The motorcycle draws level 18 s after the car passed, having travelled 180 + 30 × 6 = 360 m, which matches the car's 20 × 18 = 360 m. M1 A1 for the 12 s, M1 A1 for the two distances, M1 for the closing speed, A1 for 18 s, A1 for 360 m. Suvat applies to each phase separately but never across the join, since the acceleration changes at t = 12.

Question 15

A particle moves in a straight line with constant acceleration a. Its velocity is u at time 0 and v at time t, and its displacement in that time is s. By considering the area under the velocity-time graph, derive s = ut + ½at². Then, using v = u + at, eliminate t to derive v² = u² + 2as.

Accepted Answer

The graph is a straight line from (0, u) to (t, v) with gradient a, so v = u + at. The displacement is the area under it: a rectangle of height u and width t, plus a triangle of base t and height v − u = at. So s = ut + ½ × t × at = ut + ½at². Now substitute t = (v − u)/a: s = u(v − u)/a + ½a(v − u)²/a², so 2as = 2u(v − u) + (v − u)² = 2uv − 2u² + v² − 2uv + u² = v² − u², giving v² = u² + 2as. M1 for splitting the area into a rectangle and a triangle, A1 for s = ut + ½at², M1 for substituting t = (v − u)/a, M1 for multiplying through by 2a and expanding, A1 for the result. Checking on an instance, u = 3, a = 2, t = 4: s = 12 + 16 = 28 and v = 11, and 121 = 9 + 2 × 2 × 28 holds.

Question 16

A particle P starts at the origin O with velocity (6i − 2j) m s⁻¹ and moves with constant acceleration (−i + j) m s⁻², where i and j point east and north. Find the time at which P is moving parallel to the vector i + j, and show that at that instant P is due east of O, stating its distance from O.

Accepted Answer

v = u + at gives v = (6 − t)i + (−2 + t)j. Parallel to i + j means the two components are equal and positive: 6 − t = t − 2, so 2t = 8 and t = 4 s, when v = 2i + 2j, both components positive as required. Position: r = ut + ½at² = 4(6i − 2j) + 8(−i + j) = (24 − 8)i + (−8 + 8)j = 16i. The j component is zero and the i component positive, so P is due east of O, 16 m away. M1 for v = u + at in components, A1 for v, M1 for equating the components, A1 for t = 4, M1 for r = ut + ½at² at t = 4, A1 for 16i with the conclusion.