Calculus in Kinematics: Displacement, Velocity and Acceleration - Worksheets, Questions and Revision

5 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 5)Calculus in Kinematics: Displacement, Velocity and Acceleration revision guide

This topic is chapter 10 of IGCSE Maths Practice Book 2.

5.2 Calculus in Kinematics: Displacement, Velocity and Acceleration

EDEXCEL 4MA1 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A particle moves so that its displacement from a fixed point is s = 20t - 4t² metres after t seconds. (a) Find the velocity function and hence find the time(s) when the particle is instantaneously at rest. (b) Find the acceleration and state whether the particle is speeding up or slowing down at t = 3 seconds. Give a short reason.

(a)Find v and the time(s) when v = 0.(3)

(b)Find a and state whether the particle is speeding up or slowing down at t = 3 s, with a brief reason.(4)

(Total for Question 1 is 7 marks)

A model rocket moves vertically so that its displacement above the launch pad in metres is s = t³ - 4t² + t for time t seconds after launch. (a) Find the velocity function v = ds/dt. (b) Find the acceleration at t = 2 seconds and state, with a reason, whether the rocket is speeding up or slowing down at that instant.

(a)Find the velocity function v = ds/dt for the rocket.(2)

(b)Find the acceleration at t = 2 s and state whether the rocket is speeding up or slowing down at t = 2 s, with a brief reason.(4)

(Total for Question 2 is 6 marks)

A particle has displacement s = 0.5t³ - 3t² + 2t metres after t seconds. (a) Find expressions for velocity v and acceleration a. (b) Find the times t ≥ 0 when the particle is instantaneously at rest. (c) For each time found, state the sign of the acceleration and the immediate change in speed, with a brief reason.

(a)Find v and a.(2)

(b)Find times t ≥ 0 when v = 0.(2)

(c)For each time found, state the sign of a and whether the particle is momentarily speeding up or slowing down, with a short reason.(2)

(Total for Question 3 is 6 marks)

A motorbike's displacement along a straight road is s = -2t³ + 15t² - 24t metres after t seconds. (a) Find the acceleration a(t). (b) Find the time(s) t ≥ 0 when the acceleration is zero. (c) For one interval between these times, state whether acceleration is positive or negative and give a brief implication for the motorbike's motion.

(a)Find the acceleration function a(t).(2)

(b)Find times t ≥ 0 when acceleration is zero.(2)

(c)State the sign of acceleration for t < 2.5 and give a brief implication for the motorbike's motion in that interval.(3)

(Total for Question 4 is 7 marks)

An experimental glider has displacement s = 4t² - t³ + 6t metres from a reference point after t seconds. (a) Find v and a. (b) Find the time(s) t ≥ 0 when the glider is instantaneously at rest. (c) For each such time give the acceleration value and a one line interpretation of the motion immediately after that instant.

(a)Find expressions for v and a.(2)

(b)Find times t ≥ 0 when v = 0.(3)

(c)For each time found, give a value of a and a one line interpretation of the motion immediately after that instant.(2)

(Total for Question 5 is 7 marks)

Mark scheme · 5.2 Calculus in Kinematics: Displacement, Velocity and Acceleration

Question 1

(a) M1 differentiates s to find v = 20 - 8t or equivalent
(a) M1 sets 20 - 8t = 0 and rearranges to 8t = 20
(a) A1 t = 2.5 seconds cao
(a) Answer: v = 20 - 8t; particle at rest at t = 2.5 s
(b) M1 differentiates v to find a = dv/dt = -8
(b) M1 substitutes t = 3 into v and/or uses sign of v at t = 3 or evaluates a at t = 3 (a is constant) to compare signs
(b) A1 a = -8 m/s² cao
(b) A1 particle is slowing down at t = 3 s, with correct reason linking signs, eg velocity at t = 3 is 20 - 24 = -4 m/s and acceleration is negative so speed is increasing in the negative direction meaning the magnitude of velocity is decreasing, or simply that velocity and acceleration have same/opposite sign explained correctly
(b) Answer: a = -8 m/s²; at t = 3 s the particle is slowing down

Question 2

(a) M1 differentiates terms of s to form v, eg 3t² - 8t + 1 seen
(a) A1 v = 3t² - 8t + 1 cao
(a) Answer: v = 3t² - 8t + 1
(b) M1 differentiates v to get a = 6t - 8
(b) M1 substitutes t = 2 to find a(2) = 4
(b) A1 a = 4 m/s² cao
(b) A1 correct statement about speeding up or slowing down with reason, eg v(2) = 3(4) - 16 + 1 = -3 m/s so velocity is negative while acceleration is positive, therefore the rocket is slowing down at t = 2 s, cao
(b) Answer: a = 4 m/s²; the rocket is slowing down at t = 2 s, because v(2) = -3 m/s and a is positive so velocity and acceleration have opposite signs

Question 3

(a) M1 differentiates s to get v = 1.5t² - 6t + 2 or equivalent
(a) A1 a = 3t - 6 cao
(a) Answer: v = 1.5t² - 6t + 2; a = 3t - 6
(b) M1 sets 1.5t² - 6t + 2 = 0 and attempts to solve, e.g. multiply by 2 to 3t² - 12t + 4 = 0
(b) A1 t = (12 ± √144 - 48)/6 = (12 ± √96)/6 = (12 ± 4*√6)/6, giving t = 2 ± (2/3) √6, both values evaluated and non-negative, cao
(b) Answer: t = 2 ± (2/3) √6 seconds (both are ≥ 0)
(c) M1 evaluates a at each time correctly, recognising sign, eg a(0.367) = 3(0.367) - 6 < 0 and a(3.633) = 3(3.633) - 6 > 0
(c) A1 states correctly: at the smaller root a < 0 so since v = 0 and a < 0 the particle will start moving in the negative direction, speed will increase in the negative direction (interpreted as speeding up if velocity and acceleration same sign, here acceleration negative and velocity changes to negative so speed increases), and at the larger root a > 0 so particle will start moving in the positive direction and speed will increase in the positive direction, cao
(c) Answer: At t approx 0.367 s, a < 0 so the particle will start accelerating in the negative direction and speed will increase in that direction. At t approx 3.633 s, a > 0 so it will accelerate in the positive direction and speed will increase in the positive direction.

Question 4

(a) M1 differentiates twice, eg v = -6t² + 30t - 24 and a = -12t + 30 shown
(a) A1 a = -12t + 30 cao
(a) Answer: a = -12t + 30
(b) M1 sets -12t + 30 = 0 and solves to give t = 30/12
(b) A1 t = 2.5 seconds cao
(b) Answer: t = 2.5 s
(c) M1 evaluates a for t < 2.5, e.g. picks t = 0 or states -12t + 30 > 0 when t < 2.5
(c) M1 states sign correctly, eg a > 0 for t < 2.5
(c) A1 gives correct implication, eg motorbike is accelerating (speed increasing) while a > 0, with a brief correct reason linking sign of a to change in speed
(c) Answer: For t < 2.5 s, a > 0, so the motorbike is accelerating (its speed is increasing) in that interval.

Question 5

(a) M1 differentiates s to get v = 8t - 3t² + 6 or equivalent and a = 8 - 6t
(a) A1 v = 8t - 3t² + 6 and a = 8 - 6t cao
(a) Answer: v = 8t - 3t² + 6; a = 8 - 6t
(b) M1 sets 8t - 3t² + 6 = 0 and attempts rearrangement, e.g. 3t² - 8t - 6 = 0
(b) M1 uses quadratic formula or factorises correctly to find roots
(b) A1 t = (8 ± √64 + 72)/(6) = (8 ± √136)/6 = (8 ± 2*√34)/6, giving two real roots, numeric approximations t approx -0.215 (reject) and t approx 3.615 s, so t = 3.615 s (to 3 dp) accepted cao
(b) Answer: t approx 3.615 s (the only non-negative root)
(c) M1 evaluates a at the non-negative root correctly, e.g. a(3.615) = 8 - 6(3.615) approx -13.69
(c) A1 gives correct interpretation: a is negative so after coming to rest the glider will accelerate in the negative direction and speed will increase in that direction, cao
(c) Answer: a(3.615) approx -13.69 m/s²; since a < 0 the glider will begin to move in the negative direction and its speed in that direction will increase immediately after t = 3.615 s.

Calculus in Kinematics: Displacement, Velocity and Acceleration - Worksheets, Questions and Revision

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