Edexcel A level Maths (9MA0) · Mechanics › Kinematics
Practise Displacement, velocity and acceleration. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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A car is driven once around a closed circuit of length 1.2 km and returns to its starting point. State the displacement of the car from its starting point at the end of the lap.
Zero (0 m) m
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Which two of the following quantities are vectors? Tick (✓) two boxes.[2]
speed
displacement
distance
velocity
time
(b) A car is driven once around a closed circuit of length 1.2 km and returns to its starting point. State the displacement of the car from its starting point at the end of the lap.[1]
(c) The lap takes 80 seconds. Find the average speed of the car, giving your answer in m s−1.[2]
(d) Explain why the average velocity of the car for the lap is zero, even though its average speed is not.[1]
Show the answer and mark scheme
(a)Answer: displacement; velocity
(b)Answer: Zero (0 m) m
B1 for zero
Worked solution: The car finishes where it started, so its displacement from the starting point is zero, even though it has travelled 1.2 km.
(c)Answer: \(15\) m s−1
M1 for \(\frac{1200}{80}\)
A1 for 15 (m s−1)
Worked solution: \(\text{average speed} = \frac{\text{distance}}{\text{time}} = \frac{1200}{80} = 15\text{ m s}^{-1}\)
(d)Answer: Average velocity is displacement ÷ time, and the displacement for the lap is zero; average speed uses the distance travelled, 1200 m.
B1 for average velocity uses displacement (zero), whereas average speed uses distance travelled
Worked solution: Average velocity = displacement ÷ time = 0 ÷ 80 = 0. Average speed = distance ÷ time uses the 1200 m actually travelled.
Question 2Medium6 marks
A swimmer moves along a straight path which lies in the east–west direction. Starting from a point \(A\), the swimmer swims 240 m due east in 4 minutes, and then swims 75 m due west in 1 minute, finishing at the point \(B\).
(a) Find the average speed of the swimmer for the whole journey.[2]
(b) Find the magnitude and direction of the average velocity of the swimmer for the whole journey.[3]
(c) Explain why the average speed is greater than the magnitude of the average velocity.[1]
Show the answer and mark scheme
(a)Answer: \(1.05\) m s−1
M1 for \(\frac{240 + 75}{240 + 60}\) (distance in m and time in s)
A1 for 1.05 (m s−1)
Worked solution: Total distance \(240 + 75 = 315\) m; total time \(240 + 60 = 300\) s. \(\text{average speed} = \frac{315}{300} = 1.05\text{ m s}^{-1}\)
(b)Answer: \(0.55\) m s−1 due east m s−1
M1 for the displacement \(240 - 75 = 165\) m
M1 for \(\frac{\text{displacement}}{\text{total time}}\)
A1 for 0.55 m s−1 due east
Worked solution: The displacement of \(B\) from \(A\) is \(240 - 75 = 165\) m, i.e. 165 m due east. \(\text{average velocity} = \frac{165}{300} = 0.55\text{ m s}^{-1}\) due east.
(c)Answer: The swimmer changes direction, so the distance travelled (315 m) is greater than the magnitude of the displacement (165 m), over the same time.
B1 for the distance travelled is greater than the magnitude of the displacement because the swimmer changes direction
Worked solution: Average speed uses the total distance, 315 m, but average velocity uses the displacement, only 165 m, because part of the journey is back towards \(A\).
Question 3Hard10 marks
A particle \(P\) moves along a straight line with constant acceleration. At time \(t = 0\), \(P\) passes through the point \(O\), moving in the positive direction with speed \(u\) m s−1. At time \(t = 7\) seconds, \(P\) is at the point \(A\) and is moving in the negative direction with speed 3 m s−1. The total distance travelled by \(P\) in the interval \(0 \le t \le 7\) is 12.5 m.
(a) Explain why \(P\) is instantaneously at rest at some time during the interval \(0 \le t \le 7\).[1]
(b) Find the value of \(u\).[5]
(c) Find the acceleration of \(P\).[2]
(d) Find the displacement of \(A\) from \(O\).[2]
Show the answer and mark scheme
(a)Answer: The velocity changes from positive to negative. The acceleration is constant, so the velocity changes continuously (at a constant rate) and must be zero at some instant in between.
B1 for the velocity changes sign (positive to negative) and changes continuously, as the acceleration is constant, so it must pass through zero
Worked solution: At \(t = 0\) the velocity is positive and at \(t = 7\) it is negative. With constant acceleration the velocity changes at a constant rate, so it passes through zero: \(P\) is instantaneously at rest at that moment before moving back in the negative direction.
(b)Answer: \(u = 4\)
M1 for using the constant acceleration, e.g. \(|a| = \frac{u + 3}{7}\) or similar triangles on a velocity–time graph, to express the time at rest or the distances in terms of \(u\)
M1 for the total distance as the sum of the two distances, e.g. \(\frac{u^2}{2|a|} + \frac{3^2}{2|a|} = 12.5\)
A1 for a correct 3-term quadratic, e.g. \(7u^{2} - 25u - 12 = 0\)
dM1 for solving their quadratic
A1 for \(u = 4\) only, with the negative root rejected
Worked solution: The velocity–time graph is a straight line from \((0, u)\) to \((7, -3)\), so the magnitude of the acceleration is \(|a| = \frac{u + 3}{7}\). Distance while moving in the positive direction: \(\frac{u^2}{2|a|} = \frac{7u^2}{2(u + 3)}\); distance while moving in the negative direction: \(\frac{3^2}{2|a|} = \frac{63}{2(u + 3)}\). \(\frac{7(u^2 + 9)}{2(u + 3)} = 12.5 \Rightarrow 7(u^2 + 9) = 25(u + 3)\) \(\Rightarrow 7u^{2} - 25u - 12 = 0 \Rightarrow (u - 4)(7u + 3) = 0\), so \(u = 4\) or \(u = -\frac{3}{7}\). Since \(u\) is a speed, \(u = 4\).
(c)Answer: \(-1\) m s−2
M1 for \(\frac{v - u}{t} = \frac{-3 - 4}{7}\) using their \(u\)
A1 for \(-1\) (m s−2)
Worked solution: \(a = \frac{v - u}{t} = \frac{-3 - 4}{7} = -1\text{ m s}^{-2}\), i.e. 1 m s−2 in the negative direction.
(d)Answer: \(3.5\) m
M1 for \(s = \left(\frac{u + v}{2}\right)t = \frac{4 + (-3)}{2} \times 7\) or \(8 - 4.5\)
A1 for \(3.5\) (m)
Worked solution: \(s = \left(\frac{4 + (-3)}{2}\right) \times 7 = 3.5\) m, so \(A\) is 3.5 m from \(O\) in the positive direction (check: \(8 - 4.5 = 3.5\)).