Kinematics

A particle moves in a straight line with constant acceleration 2 m/s². Its initial velocity is 8 m/s. Find its velocity after 5 seconds.

• A10 m/s
• B8 m/s
• C18 m/s✓
• D26 m/s

Answer: Using v = u + at with u = 8, a = 2, t = 5 gives v = 8 + 2 × 5 = 18 m/s. The 10 m/s option is just at, ignoring the initial velocity; 8 m/s is the unchanged initial velocity; and 26 m/s comes from misapplying the formula as (u + t) × a = (8 + 5) × 2.

A particle accelerates uniformly in a straight line from 4 m/s to 16 m/s while travelling a distance of 60 m. Find its acceleration.

• A0.2 m/s²
• B2 m/s²✓
• C1 m/s²
• D12 m/s²

Answer: Using v² = u² + 2as: 16² = 4² + 2a(60), so 256 = 16 + 120a, giving 120a = 240 and a = 2 m/s². Treating the speed change 16 − 4 as the acceleration gives 12 m/s²; dividing that speed change by the distance gives 12/60 = 0.2 m/s²; and 1 m/s² comes from an arithmetic slip in solving 120a = 240.

A car starts from rest and accelerates uniformly to 12 m/s in 4 s, then travels at a constant 12 m/s for a further 6 s. Using the velocity–time graph, find the total distance travelled.

• A96 m✓
• B72 m
• C120 m
• D144 m

Answer: Total distance is the area under the graph: the triangle ½ × 4 × 12 = 24 m plus the rectangle 12 × 6 = 72 m, giving 24 + 72 = 96 m. Counting only the constant-speed section gives 72 m, and treating the acceleration phase as a full rectangle (12 × 4 = 48 m) rather than a triangle gives 120 m.

On a velocity–time graph a straight line rises from 5 m/s at t = 0 to 25 m/s at t = 8 s. What does the gradient of this line represent, and what is its value?

• ADisplacement, 120 m
• BAcceleration, 3.75 m/s²
• CDisplacement, 30 m
• DAcceleration, 2.5 m/s²✓

Answer: On a velocity–time graph the gradient is the acceleration: (25 − 5)/(8 − 0) = 20/8 = 2.5 m/s². Adding the velocities instead of subtracting gives 30/8 = 3.75 m/s², and any answer labelled displacement (120 m, the area under the line, or 30 m) confuses the gradient with the area.

A ball is thrown vertically upwards from ground level with a speed of 14.7 m/s. Taking g = 9.8 m/s², find the time taken to reach its maximum height.

• A0.67 s
• B3 s
• C1.5 s✓
• D1.47 s

Answer: At maximum height v = 0. Using v = u − gt: 0 = 14.7 − 9.8t, so t = 14.7/9.8 = 1.5 s. Taking g = 10 gives 14.7/10 = 1.47 s; inverting the fraction to 9.8/14.7 gives 0.67 s; and 3 s is the time for the whole flight, up and back down.

A stone is dropped from rest from the top of a tower. Taking g = 9.8 m/s², find the distance it falls in the first 3 seconds.

• A29.4 m
• B44.1 m✓
• C45 m
• D88.2 m

Answer: Using s = ut + ½gt² with u = 0: s = ½ × 9.8 × 3² = ½ × 9.8 × 9 = 44.1 m. Taking g = 10 gives 45 m; working out g × t = 9.8 × 3 = 29.4 gives the speed after 3 s, not a distance; and omitting the factor of ½ gives 9.8 × 9 = 88.2 m.

A particle moves so that its displacement from the origin is s = t³ − 6t² + 9t metres at time t seconds. Find the velocity of the particle when t = 4.

• A9 m/s✓
• B4 m/s
• C−15 m/s
• D33 m/s

Answer: Velocity is v = ds/dt = 3t² − 12t + 9. At t = 4, v = 3(16) − 12(4) + 9 = 48 − 48 + 9 = 9 m/s. The value 4 is the displacement s at t = 4, not the velocity, and 33 comes from differentiating −6t² as −6t.

A particle moving in a straight line has acceleration a = (6t − 4) m/s² at time t seconds. When t = 0 its velocity is 2 m/s. Find its velocity when t = 3.

• A14 m/s
• B27 m/s
• C21 m/s
• D17 m/s✓

Answer: Integrating, v = ∫(6t − 4) dt = 3t² − 4t + c. Using v = 2 at t = 0 gives c = 2, so v = 3t² − 4t + 2. At t = 3, v = 27 − 12 + 2 = 17 m/s. The value 14 is the acceleration 6(3) − 4 at t = 3, not the velocity, and 27 comes from keeping only the 3t² term.

A particle starts with velocity 5 m s⁻¹ and accelerates uniformly at 2 m s⁻² for 4 s. Find its velocity at the end.

• A13 m s⁻¹✓
• B36 m s⁻¹
• C20 m s⁻¹
• D8 m s⁻¹

Answer: Use v = u + at: v = 5 + (2)(4) = 13 m s⁻¹. Reaching for s = ut + ½at² gives the DISPLACEMENT, which is the wrong suvat equation for a velocity question.

A particle starts with velocity 2 m s⁻¹ and accelerates uniformly at 9.8 m s⁻² for 3 s. Find its velocity at the end.

• A6 m s⁻¹
• B50.1 m s⁻¹
• C31.4 m s⁻¹✓
• D29.4 m s⁻¹

Answer: Use v = u + at: v = 2 + (9.8)(3) = 31.4 m s⁻¹. Reaching for s = ut + ½at² gives the DISPLACEMENT, which is the wrong suvat equation for a velocity question.

A particle starts with velocity 20 m s⁻¹ and accelerates uniformly at −4 m s⁻² for 2 s. Find its velocity at the end.

• A12 m s⁻¹✓
• B32 m s⁻¹
• C40 m s⁻¹
• D−8 m s⁻¹

Answer: Use v = u + at: v = 20 + (−4)(2) = 12 m s⁻¹. Reaching for s = ut + ½at² gives the DISPLACEMENT, which is the wrong suvat equation for a velocity question.

A particle accelerates uniformly from 4 m s⁻¹ to 20 m s⁻¹ over 8 s. Find the distance travelled.

• A128
• B96✓
• C2
• D160

Answer: With uniform acceleration use s = ½(u + v)t = ½(4 + 20)(8) = 96 m. That formula needs no acceleration value, which makes it the quickest route whenever u, v and t are given.

A particle accelerates uniformly from 10 m s⁻¹ to 2 m s⁻¹ over 4 s. Find the distance travelled.

• A−32
• B24✓
• C−2
• D8

Answer: With uniform acceleration use s = ½(u + v)t = ½(10 + 2)(4) = 24 m. That formula needs no acceleration value, which makes it the quickest route whenever u, v and t are given.

A particle has displacement s = t³ − 6t² + 9t. Find its velocity at t = 2.

• A0 m s⁻¹
• B3 m s⁻¹
• C2 m s⁻¹
• D−3 m s⁻¹✓

Answer: Velocity is ds/dt = 3t² − 12t + 9. At t = 2 that is 12 − 24 + 9 = −3 m s⁻¹. The negative sign means the particle is moving back towards the origin — differentiate displacement for velocity, and again for acceleration.

On a velocity-time graph, what does the area under the graph represent?

• AThe average speed of the particle over the whole time interval
• BAcceleration
• CThe total distance travelled, regardless of the direction of motion
• DDisplacement✓

Answer: Area under a velocity-time graph gives displacement, and the GRADIENT gives acceleration. Area below the time axis counts as negative displacement, which is why displacement and distance differ when the particle turns around.

A ball is projected from level ground with speed 25 m s⁻¹ at 40° above the horizontal. Modelling the ball as a particle and taking g = 9.8 m s⁻², find its greatest height above the ground, to 3 significant figures.

• A18.7 m
• B31.9 m
• C13.2 m✓
• D26.4 m

Answer: Vertical component: 25 sin 40° = 16.07 m s⁻¹. At the top the vertical velocity is zero: 0 = 16.07² − 2(9.8)h, so h = 16.07²/19.6 = 13.2 m. Using the horizontal component 25 cos 40° gives 18.7 m, and using the full speed 25 m s⁻¹ gives 31.9 m.

A ball is thrown from a point 1.5 m above level ground with speed 20 m s⁻¹ at 30° above the horizontal. Modelling the ball as a particle and taking g = 9.8 m s⁻², find the horizontal distance travelled before the ball hits the ground, to 3 significant figures.

• A35.3 m
• B37.8 m✓
• C32.5 m
• D21.8 m

Answer: Vertically, taking up as positive: −1.5 = 10t − 4.9t², so 4.9t² − 10t − 1.5 = 0 and t = (10 + √(100 + 29.4))/9.8 = 2.181 s. Horizontal distance = 20 cos 30° × 2.181 = 37.8 m. 35.3 m ignores the 1.5 m launch height, and 32.5 m takes the landing point as 1.5 m above the launch point.

A particle is projected from level ground at an angle α above the horizontal, where tan α = 3/4, and lands 48 m from its point of projection. Taking g = 9.8 m s⁻², find its speed of projection, to 3 significant figures.

• A22.1 m s⁻¹✓
• B31.3 m s⁻¹
• C28.0 m s⁻¹
• D15.7 m s⁻¹

Answer: sin α = 0.6 and cos α = 0.8. Time of flight = 2u sin α/g = 1.2u/9.8, and range = u cos α × time = 0.96u²/9.8 = 48. So u² = 490 and u = 22.1 m s⁻¹. Dropping the factor 2 from the time of flight gives u² = 980 and 31.3 m s⁻¹.

A boat switches off its engine when moving at 15 m s⁻¹. Its deceleration due to water resistance is 0.4v m s⁻², where v m s⁻¹ is its speed. Find the time taken for its speed to fall to 5 m s⁻¹, to 3 significant figures.

• A25.0 s
• B2.75 s✓
• C0.439 s
• D4.02 s

Answer: dv/dt = −0.4v, so ∫ (1/v) dv = ∫ −0.4 dt and ln v = −0.4t + ln 15. Speed 5 gives ln(15/5) = 0.4t, so t = (ln 3)/0.4 = 2.75 s. 25.0 s treats the deceleration as a constant 0.4 m s⁻², and 0.439 s multiplies ln 3 by 0.4 instead of dividing.

A particle starts from rest and moves in a straight line with acceleration a = (2 − 0.5t) m s⁻² at time t seconds. Find the greatest speed of the particle in the interval 0 ≤ t ≤ 8.

• A8 m s⁻¹
• B2 m s⁻¹
• C6 m s⁻¹
• D4 m s⁻¹✓

Answer: v = ∫ (2 − 0.5t) dt = 2t − 0.25t², with v = 0 at t = 0. The speed is greatest when a = 0, i.e. t = 4, giving v = 8 − 4 = 4 m s⁻¹. 2 m s⁻² is the greatest acceleration, not speed; 8 m s⁻¹ uses v = 2t, ignoring the −0.5t term.

A particle moves in a straight line with velocity v = t² − 8t + 12 m s⁻¹ at time t seconds. Find the distance travelled by the particle between the two instants at which it is at rest.

• A64/3 m
• B16/3 m
• C32/3 m✓
• D8 m

Answer: v = (t − 2)(t − 6), so the particle is at rest at t = 2 and t = 6. ∫₂⁶ (t² − 8t + 12) dt = [t³/3 − 4t² + 12t]₂⁶ = 0 − 32/3 = −32/3, and v < 0 throughout, so the distance is 32/3 m. 64/3 m adds on the motion from t = 0 to t = 2, which is outside the interval.