Kinematics of Motion in a Straight Line: Mathematics 9709 (Cambridge International AS & A Level)

Syllabus 4.2 · Strand 4 Mechanics

Questions
10
Total marks
52
Tier mix
10 Core

0 of 10 questions completed

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Syllabus coverage

  • 4.2 10 questions

Kinematics (syllabus ref 4.2), restricted here to motion along a single straight line, opens with a careful vocabulary distinction: distance and speed are scalars (size only), while displacement, velocity and acceleration are vectors (size and direction), so a change in direction can make displacement smaller even while distance travelled keeps growing. Two graphs summarise this motion: on a displacement–time graph, the gradient at any point gives velocity; on a velocity–time graph, the gradient gives acceleration and the area under the curve gives displacement.

Where velocity or acceleration is given as a function of time, calculus connects the three quantities directly: differentiating displacement gives velocity, differentiating velocity gives acceleration, and integrating in the opposite direction recovers displacement or velocity (using P1 calculus techniques). For constant acceleration, the standard “suvat” formulae (such as v=u+atv=u+at, s=ut+12at2s=ut+\tfrac12at^2, and v2=u2+2asv^2=u^2+2as) let you solve for any missing quantity directly, and some questions set up two such equations at once, e.g. comparing the motion of two separate particles that start at different times or positions.

Original worked examples below cover graphical, calculus-based and constant-acceleration approaches to this topic.

Question 1

Multiple choice AS 1 mark

A go-kart passes a marker post on a straight track with a velocity of 3 m s13\text{ m s}^{-1} and then accelerates uniformly at 1.5 m s21.5\text{ m s}^{-2}.

Find the velocity of the go-kart 66 seconds after it passes the marker post.

Question 2

Structured AS 8 marks

A cyclist rides along a straight road for 4040 seconds, starting from rest at a fixed point OO. The velocity–time graph of her motion consists of three straight-line stages:

  • Stage 1 (0t100 \le t \le 10): the velocity increases uniformly from 00 to 8 m s18\text{ m s}^{-1}.
  • Stage 2 (10t3010 \le t \le 30): the velocity stays constant at 8 m s18\text{ m s}^{-1}.
  • Stage 3 (30t4030 \le t \le 40): the velocity decreases uniformly from 8 m s18\text{ m s}^{-1} back to 00.

(a) Find the acceleration of the cyclist during Stage 1 and during Stage 3. [2]

(b) By considering the area under each stage of the velocity–time graph, find the total distance travelled by the cyclist during the 4040 seconds. [4]

(c) Find the average velocity of the cyclist over the whole 4040 seconds. [2]

Question 3

Structured AS 8 marks

A delivery robot moves along a straight corridor, starting from rest at a fixed point OO. Its acceleration a m s2a\text{ m s}^{-2}, at time tt seconds after leaving OO, is given by a=63tfor 0t4.a = 6 - 3t \quad \text{for } 0 \le t \le 4.

(a) Show that the velocity of the robot at time tt is given by v=6t1.5t2v = 6t - 1.5t^2. [3]

(b) Find the maximum velocity attained by the robot, and the value of tt at which it occurs. Justify why this gives a maximum rather than a minimum. [3]

(c) Find the total distance travelled by the robot during 0t40 \le t \le 4. [2]

Question 4

Structured AS 8 marks

A stone is thrown vertically upward from ground level with an initial speed of 25 m s125\text{ m s}^{-1}. The stone is modelled as a particle moving in a straight vertical line, and air resistance is ignored. Take g=10 m s2g = 10\text{ m s}^{-2}, and take the upward direction as positive.

(a) Find the greatest height above the ground reached by the stone. [3]

(b) Find the total time taken for the stone to return to the ground. [3]

(c) Find the speed at which the stone hits the ground. [2]

Question 5

Multiple choice AS 1 mark

A particle moves along a straight line. Its displacement ss metres from a fixed point OO, tt seconds after it starts to move, has a displacement–time graph made of two straight-line stages:

  • For 0t40 \le t \le 4, ss increases uniformly from 00 m to 2020 m.
  • For 4t104 \le t \le 10, ss decreases uniformly from 2020 m to 55 m.

Find the velocity of the particle during the interval 4t104 \le t \le 10.

Question 6

Structured AS 8 marks

A train travels along a straight, horizontal section of track at a constant velocity of 24 m s124\text{ m s}^{-1}. As it approaches a station, the driver applies the brakes, giving the train a constant deceleration. The train comes to rest after travelling a further 180180 m.

(a) Find the deceleration of the train. [3]

(b) Find the time taken for the train to come to rest after the brakes are applied. [3]

(c) Find the average velocity of the train while it is decelerating, and use it to verify your answer to part (b). [2]

Question 7

Multiple choice AS 1 mark

A skateboarder starts from rest at the top of a straight ramp and accelerates uniformly down the ramp at 0.4 m s20.4\text{ m s}^{-2}.

Find the distance she has travelled after 55 seconds.

Question 8

Multiple choice AS 1 mark

A particle moves along a straight line. Its velocity–time graph consists of two straight-line stages:

  • For 0t40 \le t \le 4, the velocity decreases uniformly from 8 m s18\text{ m s}^{-1} to 00.
  • For 4t64 \le t \le 6, the velocity continues to decrease uniformly from 00 to 6 m s1-6\text{ m s}^{-1} (the particle is now moving back towards its starting point).

Find the total displacement of the particle from its starting point at t=6t=6.

Question 9

Structured AS 8 marks

A particle PP moves along a straight line so that its displacement ss metres from a fixed point OO, at time tt seconds (t0t\ge0), is given by s=t36t2+9t.s = t^3 - 6t^2 + 9t.

(a) Find expressions for the velocity vv and the acceleration aa of PP at time tt. [3]

(b) Find the values of tt at which PP is instantaneously at rest. [3]

(c) Find the acceleration of PP at each of the values of tt found in part (b), and use these to state, in each case, whether PP is about to change direction. [2]

Question 10

Structured AS 8 marks

Car AA starts from rest at a fixed point OO on a straight road and moves with constant acceleration 2 m s22\text{ m s}^{-2}. Car BB travels along the same road in the same direction at a constant velocity of 20 m s120\text{ m s}^{-1}, and passes through OO exactly 55 seconds after Car AA sets off. Let tt be the time in seconds measured from the instant Car AA sets off.

(a) Write down expressions, in terms of tt, for the displacement from OO of Car AA, and of Car BB (valid for t5t\ge5). [3]

(b) Show that the times at which the two cars are at the same distance from OO satisfy (t10)2=0(t-10)^2=0, and hence find this value of tt. [3]

(c) State what a repeated root tells you about the motion of the two cars, and find the distance from OO at which the two cars are momentarily at the same position. [2]