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. Option A uses only a × t, option B forgets the acceleration, and D wrongly adds u + at² = 8 + 2 × 25.

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. Option D wrongly uses (16 − 4) as the acceleration; A and C come from arithmetic slips.

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. Option B counts only the constant-speed section.

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. Option B divides 30 by 8; options A and C confuse gradient (acceleration) with area (displacement).

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. Option D wrongly uses g = 10, and A inverts the fraction.

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. Option C uses g = 10, option A uses g × t, and D omits the factor of ½.

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. Option C evaluates the displacement instead, and D mishandles the middle term.

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. Option A forgets the constant; B omits the −4t term.