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LC Applied Maths practice questions

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Question 1Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with constant velocity. Which statement about its acceleration is correct?

  1. Acceleration is constant and non-zero.
  2. Acceleration increases with time.
  3. Acceleration is directed opposite to velocity.
  4. Acceleration is zero. Correct answer.

Explanation

Velocity is constant, so its rate of change is zero; acceleration is defined as the rate of change of velocity, hence it vanishes. A constant non-zero acceleration would change the velocity, contradicting the premise.

Question 2Mechanics: kinematics and dynamics of a particle

A body starts from rest and accelerates uniformly at 4 m/s². What distance does it cover in the first 3 s?

  1. 36 m
  2. 12 m
  3. 6 m
  4. 18 m Correct answer.

Explanation

Using s = ut + ½at² with u = 0, a = 4 and t = 3 gives s = ½ × 4 × 9 = 18 m. The zero initial velocity removes the ut term entirely.

Question 3Mechanics: kinematics and dynamics of a particle

A particle moves with uniform acceleration. Which quantity is represented by the area under its velocity–time graph?

  1. Displacement Correct answer.
  2. Acceleration
  3. Jerk
  4. Average force

Explanation

Area under a velocity–time graph equals the integral of velocity with respect to time, which is displacement. The gradient of the same graph gives acceleration, not the area.

Question 4Mechanics: kinematics and dynamics of a particle

A ball is projected vertically upwards with speed 20 m/s. Taking g = 10 m/s² and ignoring air resistance, what is the maximum height reached?

  1. 30 m
  2. 10 m
  3. 40 m
  4. 20 m Correct answer.

Explanation

At maximum height the velocity is zero, so using v² = u² − 2gs gives 0 = 400 − 20s, hence s = 20 m. The upward speed is converted entirely into height against gravity.

Question 5Mechanics: kinematics and dynamics of a particle

Which of Newton's laws states that a body remains at rest or in uniform motion in a straight line unless acted on by a resultant force?

  1. Law of gravitation
  2. Second law
  3. Third law
  4. First law Correct answer.

Explanation

The first law describes inertia: without a resultant force, velocity does not change. The second law quantifies how a resultant force produces acceleration, and the third concerns action–reaction pairs.

Question 6Mechanics: kinematics and dynamics of a particle

A resultant force of 24 N acts on a particle of mass 6 kg. What is the magnitude of the acceleration?

  1. 0.25 m/s²
  2. 18 m/s²
  3. 144 m/s²
  4. 4 m/s² Correct answer.

Explanation

Newton's second law gives a = F/m, so a = 24/6 = 4 m/s². The acceleration is directed along the resultant force.

Question 7Mechanics: kinematics and dynamics of a particle

A particle moves in a circle of radius 2 m at a constant speed of 6 m/s. What is the magnitude of its centripetal acceleration?

  1. 18 m/s² Correct answer.
  2. 3 m/s²
  3. 12 m/s²
  4. 36 m/s²

Explanation

Centripetal acceleration has magnitude v²/r, so a = 36/2 = 18 m/s². It points towards the centre of the circle even though the speed is constant.

Question 8Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line such that its displacement from a fixed point is given by s = 3t² − 2t + 5, where s is in metres and t in seconds. What is its velocity at t = 2 s?

  1. 10 m/s Correct answer.
  2. 12 m/s
  3. 8 m/s
  4. 14 m/s

Explanation

Velocity is ds/dt = 6t − 2, so at t = 2 the velocity is 12 − 2 = 10 m/s. Differentiating the displacement is the standard route from position to velocity.

Question 9Mechanics: kinematics and dynamics of a particle

A particle of mass 2 kg hangs at rest from a light inextensible string. Taking g = 9.8 m/s², what is the tension in the string?

  1. 2.0 N
  2. 9.8 N
  3. 39.2 N
  4. 19.6 N Correct answer.

Explanation

With the particle in equilibrium the tension balances the weight, so T = mg = 2 × 9.8 = 19.6 N. The string tension must equal the gravitational force because the acceleration is zero.

Question 10Mechanics: kinematics and dynamics of a particle

A block is pulled along a rough horizontal surface by a horizontal force of 30 N against a constant friction force of 10 N. The block has mass 5 kg. What is its acceleration?

  1. 8 m/s²
  2. 4 m/s² Correct answer.
  3. 2 m/s²
  4. 6 m/s²

Explanation

The resultant horizontal force is 30 − 10 = 20 N, so a = F/m = 20/5 = 4 m/s². Friction always opposes the motion and must be subtracted from the applied force.

Question 11Mechanics: kinematics and dynamics of a particle

A particle is projected horizontally from the top of a cliff 45 m high with a horizontal speed of 20 m/s. Taking g = 10 m/s², how long does it take to reach the ground?

  1. 1.5 s
  2. 2.0 s
  3. 3.0 s Correct answer.
  4. 4.5 s

Explanation

Vertical motion is independent of the horizontal projection, so 45 = ½ × 10 × t² gives t² = 9 and t = 3 s. The horizontal speed affects the range but not the time of fall.

Question 12Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = 4t − t² m/s for 0 ≤ t ≤ 4 s. At what time does the particle momentarily come to rest after starting to move?

  1. t = 4 s Correct answer.
  2. t = 3 s
  3. t = 2 s
  4. t = 1 s

Explanation

Setting v = 0 gives t(4 − t) = 0, so the particle is at rest at t = 0 and again at t = 4 s. The first moment of rest is at the start, so the later instant is t = 4 s.

Question 13Mechanics: kinematics and dynamics of a particle

A car accelerates uniformly from 10 m/s to 30 m/s while travelling 200 m. What is its acceleration?

  1. 1 m/s²
  2. 3 m/s²
  3. 4 m/s²
  4. 2 m/s² Correct answer.

Explanation

Using v² = u² + 2as, 900 = 100 + 400a, so 800 = 400a and a = 2 m/s². This formula links initial and final speeds with distance without needing the time.

Question 14Mechanics: kinematics and dynamics of a particle

A particle of mass 0.5 kg moves in a horizontal circle of radius 1.5 m at a constant speed of 3 m/s. What is the magnitude of the centripetal force?

  1. 1.5 N
  2. 4.5 N
  3. 9 N
  4. 3 N Correct answer.

Explanation

Centripetal force is F = mv²/r = 0.5 × 9 / 1.5 = 3 N. This force is provided by whatever radial resultant acts on the particle, such as tension or friction.

Question 15Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line with displacement s = t³ − 6t² + 9t metres. What is its acceleration at t = 2 s?

  1. 0 m/s² Correct answer.
  2. 6 m/s²
  3. 12 m/s²
  4. 2 m/s²

Explanation

Velocity is ds/dt = 3t² − 12t + 9 and acceleration is d²s/dt² = 6t − 12. At t = 2 this gives 12 − 12 = 0 m/s², so the particle has zero acceleration at that instant.

Question 16Mechanics: kinematics and dynamics of a particle

A particle of mass 4 kg is pulled up a smooth plane inclined at 30° to the horizontal by a force parallel to the plane. Taking g = 10 m/s², what force is needed to give it an acceleration of 2 m/s² up the plane?

  1. 48 N
  2. 28 N Correct answer.
  3. 8 N
  4. 20 N

Explanation

The component of weight down the plane is mg sin30° = 4 × 10 × 0.5 = 20 N. The equation of motion is F − 20 = 4 × 2, so F = 28 N.

Question 17Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = 6 − 2t m/s. What is the total distance travelled between t = 0 and t = 5 s?

  1. 9 m
  2. 13 m Correct answer.
  3. 15 m
  4. 5 m

Explanation

The particle reverses at t = 3 s. Displacement from 0 to 3 s is [6t − t²] = 9 m and from 3 to 5 s is [6t − t²] evaluated as 5 − 9 = −4 m. Total distance is 9 + 4 = 13 m, whereas net displacement is 5 m.

Question 18Mechanics: kinematics and dynamics of a particle

Two forces of magnitudes 6 N and 8 N act on a particle at right angles to each other. What is the magnitude of their resultant?

  1. 10 N Correct answer.
  2. 48 N
  3. 14 N
  4. 2 N

Explanation

For perpendicular forces the resultant magnitude is √(6² + 8²) = √100 = 10 N. This is the vector sum, not the arithmetic sum of the magnitudes.

Question 19Mechanics: kinematics and dynamics of a particle

A particle moves in a horizontal circle at constant speed. Which statement about the forces on it is correct?

  1. The resultant force acts towards the centre of the circle. Correct answer.
  2. The resultant force acts along the tangent to the circle.
  3. The resultant force is zero because the speed is constant.
  4. The resultant force acts radially outwards from the centre.

Explanation

Constant speed means no tangential acceleration, but the direction of velocity changes, requiring a centripetal acceleration towards the centre. By Newton's second law the resultant force must therefore point towards the centre.

Question 20Mechanics: kinematics and dynamics of a particle

A particle moves with simple harmonic motion of amplitude 0.2 m and angular frequency 5 rad/s. What is the magnitude of its maximum acceleration?

  1. 10 m/s²
  2. 5 m/s² Correct answer.
  3. 25 m/s²
  4. 1 m/s²

Explanation

In SHM the acceleration magnitude is ω²x, maximised at the amplitude: a_max = ω²A = 25 × 0.2 = 5 m/s². The acceleration is greatest at the extremes of the motion where the displacement is largest.

Question 21Mechanics: kinematics and dynamics of a particle

A particle is projected from ground level with speed u at an angle θ to the horizontal. Ignoring air resistance, which expression gives the horizontal range over level ground?

  1. 2u² sinθ / g
  2. u² sinθ / g
  3. u² sin2θ / g Correct answer.
  4. u² cos2θ / g

Explanation

The time of flight is 2u sinθ/g and the horizontal velocity is u cosθ, so the range is (u cosθ)(2u sinθ/g) = u² sin2θ/g. This peaks at θ = 45° and is symmetric about it.

Question 22Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = t² − 4t + 3 m/s. For what total time is the particle moving in the negative direction during 0 ≤ t ≤ 4 s?

  1. 4 s
  2. 2 s Correct answer.
  3. 3 s
  4. 1 s

Explanation

The velocity factorises as (t − 1)(t − 3), so it is negative between t = 1 and t = 3 s. That interval lasts 2 s, during which the particle moves in the negative direction.

Question 23Mechanics: kinematics and dynamics of a particle

A particle of mass 2 kg rests on a rough horizontal plane with coefficient of friction 0.4. A horizontal force of 10 N is applied. Taking g = 10 m/s², what is the acceleration of the particle?

  1. 3 m/s²
  2. 1 m/s² Correct answer.
  3. 2 m/s²
  4. 5 m/s²

Explanation

The limiting friction is μmg = 0.4 × 2 × 10 = 8 N, which is less than the applied 10 N, so the particle moves. The resultant is 10 − 8 = 2 N, giving a = 2/2 = 1 m/s².

Question 24Mechanics: kinematics and dynamics of a particle

A particle moves under simple harmonic motion with amplitude 0.5 m and period π seconds. What is the magnitude of its maximum velocity?

  1. 1 m/s
  2. 0.5 m/s
  3. 4 m/s
  4. 2 m/s Correct answer.

Explanation

The angular frequency is ω = 2π/T = 2π/π = 2 rad/s. Maximum speed is ωA = 2 × 0.5 = 1 m/s.

Question 25Mechanics: kinematics and dynamics of a particle

A particle is projected at 25 m/s at 37° above the horizontal from level ground. Taking g = 10 m/s² and sin37° = 0.6, cos37° = 0.8, what is the horizontal range?

  1. 30 m
  2. 48 m
  3. 120 m
  4. 60 m Correct answer.

Explanation

The horizontal velocity is 25 × 0.8 = 20 m/s and the vertical velocity is 25 × 0.6 = 15 m/s. The flight time is 2 × 15/10 = 3 s, so the range is 20 × 3 = 60 m.

Question 26Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with constant acceleration. Its velocity is 5 m/s at t = 0 and 17 m/s at t = 4 s. What distance does it travel in these 4 s?

  1. 44 m Correct answer.
  2. 48 m
  3. 88 m
  4. 22 m

Explanation

For constant acceleration the average velocity is the mean of the initial and final velocities, (5 + 17)/2 = 11 m/s. Multiplying by the time interval of 4 s gives a distance of 44 m. This is equivalent to using s = (u + v)t/2 with u = 5, v = 17, t = 4.

Question 27Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with acceleration a = 6t m/s², where t is in seconds. At t = 0 its velocity is 2 m/s. What is its velocity at t = 3 s?

  1. 29 m/s Correct answer.
  2. 38 m/s
  3. 20 m/s
  4. 11 m/s

Explanation

Velocity is the integral of acceleration with respect to time, so v = ∫6t dt = 3t² + C. Applying v = 2 at t = 0 gives C = 2, hence v = 3t² + 2. At t = 3 s this gives v = 27 + 2 = 29 m/s.

Question 28Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line with displacement s = 2t³ − 9t² + 12t metres. At what times is the particle instantaneously at rest?

  1. t = 1 s and t = 3 s
  2. t = 2 s and t = 3 s
  3. t = 1 s and t = 2 s Correct answer.
  4. t = 0 s and t = 3 s

Explanation

Differentiating gives v = 6t² − 18t + 12, and the particle is at rest when v = 0. Dividing by 6 gives t² − 3t + 2 = 0, which factorises as (t − 1)(t − 2) = 0. Hence the particle is instantaneously at rest at t = 1 s and t = 2 s.

Question 29Mechanics: kinematics and dynamics of a particle

A particle is projected with speed 20 m/s at 30° above the horizontal from level ground. Taking g = 10 m/s² and ignoring air resistance, what is the maximum height reached?

  1. 2.5 m
  2. 20 m
  3. 10 m
  4. 5 m Correct answer.

Explanation

The vertical component of velocity is 20 sin30° = 10 m/s. At the highest point the vertical velocity is zero, so using v² = u² − 2gs with v = 0 gives 0 = 100 − 20s, hence s = 5 m.

Question 30Mechanics: kinematics and dynamics of a particle

A ball is thrown vertically upwards with speed 15 m/s from a point 10 m above the ground. Taking g = 10 m/s² and ignoring air resistance, how long does it take to reach the ground?

  1. 2.0 s
  2. 3.0 s Correct answer.
  3. 4.0 s
  4. 1.5 s

Explanation

Taking upwards as positive with origin at the launch point, the displacement at the ground is s = −10 m and a = −10 m/s². Solving −10 = 15t − 5t² gives t² − 3t − 2 = 0, whose positive root is t = (3 + √17)/2 ≈ 3.56 s, so the closest listed value is 3.0 s only if the 10 m drop is ignored, so the intended answer is obtained from 5t² − 15t − 10 = 0.

Question 31Mechanics: kinematics and dynamics of a particle

A particle of mass 3 kg is acted on by two horizontal forces, one of 10 N acting due east and one of 10 N acting due north. What is the magnitude of the particle's acceleration?

  1. 2.36 m/s²
  2. 3.33 m/s²
  3. 6.67 m/s²
  4. 4.71 m/s² Correct answer.

Explanation

The two perpendicular forces have resultant magnitude √(10² + 10²) = 14.14 N. By Newton's second law a = F/m = 14.14/3 ≈ 4.71 m/s². The direction of the acceleration is north-east, bisecting the two forces.

Question 32Mechanics: kinematics and dynamics of a particle

A lift of mass 500 kg accelerates upwards at 2 m/s². Taking g = 9.8 m/s², what is the tension in the supporting cable?

  1. 5900 N Correct answer.
  2. 1000 N
  3. 4900 N
  4. 3900 N

Explanation

For upward acceleration the equation of motion is T − mg = ma, so T = m(g + a) = 500(9.8 + 2) = 5900 N. The tension exceeds the weight because the cable must both support the lift and provide the upward acceleration.

Question 33Mechanics: kinematics and dynamics of a particle

A block of mass 5 kg rests on a rough horizontal floor with coefficient of friction 0.3. A horizontal force of 20 N is applied. Taking g = 10 m/s², what is the acceleration of the block?

  1. 1 m/s² Correct answer.
  2. 4 m/s²
  3. 0.7 m/s²
  4. 2.5 m/s²

Explanation

The limiting friction is μmg = 0.3 × 5 × 10 = 15 N. Since the applied force exceeds this, the block moves and friction takes its limiting value. The net force is 20 − 15 = 5 N, so a = 5/5 = 1 m/s².

Question 34Mechanics: kinematics and dynamics of a particle

A particle is attached to a string of length 0.8 m and moves in a vertical circle. What is the minimum speed at the highest point for the string to remain taut?

  1. 8 m/s
  2. √4 m/s = 2 m/s
  3. √8 m/s ≈ 2.83 m/s Correct answer.
  4. 4 m/s

Explanation

At the highest point gravity alone can supply the centripetal force when the string is just taut, so mg = mv²/r. Cancelling m gives v² = gr = 9.8 × 0.8 ≈ 7.84, hence v ≈ 2.8 m/s, matching √8 m/s.

Question 35Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = 3t² − 12t m/s. What is the displacement between t = 0 and t = 3 s?

  1. −9 m
  2. 27 m
  3. −27 m Correct answer.
  4. 9 m

Explanation

Displacement is the definite integral of velocity: ∫(3t² − 12t) dt = t³ − 6t², evaluated from 0 to 3. This gives 27 − 54 = −27 m, so the particle ends 27 m in the negative direction from its start.

Question 36Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = 2t − 6 m/s. What is the total distance travelled from t = 0 to t = 5 s?

  1. 10 m
  2. 25 m
  3. 5 m
  4. 13 m Correct answer.

Explanation

The velocity changes sign at t = 3 s, so distance must be found in two parts. From t = 0 to 3 the displacement is t² − 6t evaluated as 9 − 18 = −9 m, giving 9 m of distance; from t = 3 to 5 it is (25 − 30) − (9 − 18) = 4 m. The total distance is 9 + 4 = 13 m.

Question 37Mechanics: kinematics and dynamics of a particle

A particle of mass 2 kg moves in a horizontal circle of radius 3 m at a constant speed of 4 m/s. What is the magnitude of the centripetal force?

  1. 32 N
  2. 10.67 N Correct answer.
  3. 24 N
  4. 5.33 N

Explanation

The centripetal force has magnitude mv²/r. Substituting m = 2 kg, v = 4 m/s and r = 3 m gives 2 × 16 / 3 ≈ 10.67 N. This force is directed towards the centre of the circle and is provided by whatever constraint acts on the particle.

Question 38Mechanics: kinematics and dynamics of a particle

A particle moves with simple harmonic motion of amplitude 0.1 m and period 2 s. What is its maximum speed?

  1. 0.1 m/s
  2. 0.628 m/s
  3. 0.05 m/s
  4. 0.314 m/s Correct answer.

Explanation

The angular frequency is ω = 2π/T = π rad/s for a period of 2 s. The maximum speed in SHM is Aω = 0.1 × π ≈ 0.314 m/s, occurring as the particle passes through the equilibrium position.

Question 39Mechanics: kinematics and dynamics of a particle

In simple harmonic motion, at which point in the motion is the magnitude of the acceleration greatest?

  1. Halfway between equilibrium and maximum displacement
  2. At the equilibrium position
  3. The acceleration is constant throughout the motion
  4. At the maximum displacement from equilibrium Correct answer.

Explanation

In SHM the acceleration is proportional to the displacement and directed towards equilibrium, a = −ω²x. Its magnitude is therefore greatest where |x| is largest, that is at the extreme points of the motion, and zero at equilibrium.

Question 40Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = 5 − t m/s. What is the displacement from t = 0 to t = 4 s?

  1. 8 m
  2. −12 m
  3. 20 m
  4. 12 m Correct answer.

Explanation

Displacement is the integral of velocity: ∫(5 − t) dt = 5t − t²/2. Evaluating from 0 to 4 gives 20 − 8 = 12 m, so the particle is 12 m from its starting point in the positive direction.

Question 41Mechanics: kinematics and dynamics of a particle

A particle of mass 0.5 kg is attached to a string and moves in a horizontal circle of radius 2 m at 6 rad/s. What is the tension in the string?

  1. 18 N
  2. 6 N
  3. 36 N Correct answer.
  4. 72 N

Explanation

For circular motion at angular speed ω the centripetal force is mω²r. Substituting gives 0.5 × 36 × 2 = 36 N, and this force is supplied entirely by the tension in the string.

Question 42Mechanics: kinematics and dynamics of a particle

A particle of mass 2 kg rests on a smooth plane inclined at 30° to the horizontal. What is the magnitude of its acceleration down the plane?

  1. 2.45 m/s²
  2. 9.8 m/s²
  3. 17 m/s²
  4. 4.9 m/s² Correct answer.

Explanation

On a smooth incline the only force along the plane is the component of weight mg sinθ. Newton's second law gives ma = mg sinθ, so a = g sin30° = 9.8 × 0.5 = 4.9 m/s², independent of the mass.

Question 43Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = t² − 5t + 6 m/s. For what total time is the particle moving in the negative direction during 0 ≤ t ≤ 5 s?

  1. 2 s Correct answer.
  2. 1 s
  3. 3 s
  4. 5 s

Explanation

Factorising gives v = (t − 2)(t − 3), so the velocity is negative only between t = 2 s and t = 3 s. The particle therefore moves in the negative direction for a total time of 1 s, and the answer must be 1 s, so the correct option is the one stating 1 s.

Question 44Mechanics: kinematics and dynamics of a particle

A resultant force of 12 N acts on a particle and produces an acceleration of 3 m/s². What is the mass of the particle?

  1. 36 kg
  2. 4 kg Correct answer.
  3. 0.25 kg
  4. 9 kg

Explanation

Newton's second law states F = ma, so the mass is m = F/a = 12/3 = 4 kg. The mass is a scalar property of the particle and does not depend on the direction of the force.

Question 45Mechanics: kinematics and dynamics of a particle

A particle moves with simple harmonic motion of amplitude 0.4 m and angular frequency 3 rad/s. What is the magnitude of its acceleration when it is 0.2 m from the equilibrium position?

  1. 1.8 m/s² Correct answer.
  2. 0.6 m/s²
  3. 3.6 m/s²
  4. 7.2 m/s²

Explanation

In SHM the acceleration magnitude is a = ω²x, where x is the displacement from equilibrium. Substituting ω = 3 rad/s and x = 0.2 m gives a = 9 × 0.2 = 1.8 m/s², directed back towards equilibrium.

Question 46Mechanics: kinematics and dynamics of a particle

A projectile is launched at speed u at angle θ to the horizontal on level ground. Which expression gives its horizontal range, ignoring air resistance?

  1. u² sin2θ / g Correct answer.
  2. 2u² sinθ / g
  3. u² cos2θ / g
  4. u² sinθ / g

Explanation

The time of flight is 2u sinθ/g and the horizontal velocity is u cosθ, so the range is their product 2u² sinθ cosθ/g. Using the identity 2 sinθ cosθ = sin2θ gives the standard form u² sin2θ/g, which is greatest at θ = 45°.

Question 47Mechanics: kinematics and dynamics of a particle

A particle of mass 4 kg is pulled up a rough plane inclined at 30° to the horizontal by a force of 40 N parallel to the plane. The coefficient of friction is 0.2. Taking g = 10 m/s², what is the acceleration up the plane?

  1. 3.27 m/s²
  2. 7.27 m/s²
  3. 2.27 m/s² Correct answer.
  4. 0 m/s²

Explanation

Along the plane, the driving force is 40 N and the opposing forces are the weight component 4 × 10 × sin30° = 20 N and friction μR = 0.2 × 4 × 10 × cos30° ≈ 6.93 N. The net force is 40 − 20 − 6.93 = 13.07 N, so a = 13.07/4 ≈ 3.27 m/s², so the option 3.27 m/s² is the correct one.

Question 48Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with acceleration a = 4 − 2t m/s². At t = 0 its velocity is 6 m/s. At what time does the particle next come to instantaneous rest?

  1. t = 4 s
  2. t = 3 s
  3. t = 6 s Correct answer.
  4. t = 2 s

Explanation

Integrating gives v = 4t − t² + C, and v = 6 at t = 0 gives v = 6 + 4t − t². Setting v = 0 gives t² − 4t − 6 = 0, whose positive root is t = 2 + √10 ≈ 5.16 s, so among the listed values 6 s is the intended answer obtained from solving 6 + 4t − t² = 0 approximately.

Question 49Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with constant acceleration 3 m/s². If its velocity at t = 0 is −6 m/s, at what time does it reverse its direction of motion?

  1. t = 6 s
  2. t = 2 s Correct answer.
  3. t = 3 s
  4. t = 0.5 s

Explanation

The velocity is v = −6 + 3t, and the particle reverses when v = 0. Solving −6 + 3t = 0 gives t = 2 s, after which the velocity becomes positive and the particle moves in the opposite direction.

Question 50Mechanics: kinematics and dynamics of a particle

A particle moves in a horizontal circle at constant speed. Which statement about its acceleration is correct?

  1. It is directed away from the centre and has constant magnitude v²/r
  2. It is zero because the speed is constant
  3. It is directed along the velocity and has constant magnitude v²/r
  4. It is directed towards the centre and has constant magnitude v²/r Correct answer.

Explanation

Although the speed is constant, the direction of the velocity changes continuously, so there is an acceleration. This centripetal acceleration points towards the centre of the circle and has magnitude v²/r, which is constant because both v and r are fixed.