Edexcel A level Maths (9MA0) · Mechanics › Kinematics
Practise Constant acceleration (suvat). 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.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A car moves along a straight horizontal road with constant acceleration. It passes the point \(A\) with speed 5 m s−1 and, 8 seconds later, passes the point \(B\) with speed 11.4 m s−1.
(a) Find the acceleration of the car.[2]
(b) Find the distance \(AB\).[2]
Show the answer and mark scheme
(a)Answer: \(0.8\) m s−2
M1 for \(11.4 = 5 + 8a\)
A1 for 0.8 (m s−2)
Worked solution: \(v = u + at \Rightarrow 11.4 = 5 + 8a \Rightarrow a = 0.8\text{ m s}^{-2}\)
(b)Answer: \(65.6\) m
M1 for \(s = \frac{(5 + 11.4)}{2} \times 8\) or another complete method
A1 for 65.6 (m)
Worked solution: \(s = \left(\frac{u + v}{2}\right)t = \frac{5 + 11.4}{2} \times 8 = 65.6\) m
Question 2Medium7 marks
A particle \(P\) moves along a straight line with constant acceleration. At time \(t = 0\), \(P\) passes through the point \(O\) with velocity 8 m s−1. The acceleration of \(P\) is 4 m s−2 in the direction opposite to its initial velocity. The point \(A\) lies on the line, 6 m from \(O\) in the direction of the initial velocity of \(P\).
(a) Find the times at which \(P\) is at \(A\).[4]
(b) Explain why there are two answers to part (a).[1]
(c) Find the velocity of \(P\) at the second time it passes \(A\).[2]
Show the answer and mark scheme
(a)Answer: \(t = 1\) and \(t = 3\)
M1 for \(6 = 8t - \tfrac{1}{2} \times 4t^2\)
A1 for a correct 3-term quadratic, e.g. \(2t^{2} - 8t + 6 = 0\)
(b)Answer: \(P\) passes through \(A\) moving away from \(O\), comes instantaneously to rest beyond \(A\), and then passes through \(A\) again on its way back.
B1 for \(P\) passes \(A\) on the way out, stops, and passes \(A\) again moving back towards \(O\)
Worked solution: The acceleration opposes the motion, so \(P\) slows down, stops at a point beyond \(A\) and then returns, passing \(A\) a second time.
(c)Answer: \(-4\) m s−1
M1 for \(v = 8 - 4 \times 3\)
A1 for \(-4\) (or 4 m s−1 towards \(O\))
Worked solution: \(v = u + at = 8 - 4 \times 3 = -4\text{ m s}^{-1}\), so \(P\) is moving towards \(O\) with speed 4 m s−1.
Question 3Hard9 marks
A student drops a stone from rest at the top of a well and hears the splash 2.5 seconds later. Take \(g = 9.8\) m s−2.
(a) In Model 1, the stone falls freely under gravity and the sound of the splash is heard instantly. Use Model 1 to find the depth of the well.[2]
(b) Model 2 also takes into account the time taken for the sound to travel up the well, at a constant speed of 340 m s−1. Show that, if the stone falls for \(T\) seconds, then \(4.9T^2 = 340(2.5 - T)\).[2]
(c) Hence find the depth of the well predicted by Model 2.[3]
(d) Both models ignore air resistance. State, with a reason, whether including air resistance would make the predicted depth greater or smaller.[2]
Show the answer and mark scheme
(a)Answer: 30.6 m
M1 for \(s = \frac{1}{2} \times 9.8 \times 2.5^2\)
A1 for 30.6 (m) (accept 31)
Worked solution: \(s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(9.8)(2.5)^2 = 30.6\) m.
(b)Answer: The depth is \(4.9T^2\) (falling) and also \(340(2.5 - T)\) (the sound travels for the remaining \(2.5 - T\) seconds).
M1 for the depth \(= \frac{1}{2}(9.8)T^2 = 4.9T^2\) and the sound takes \(2.5 - T\) seconds
A1* for equating the two expressions for the depth
Worked solution: The stone falls \(\frac{1}{2}(9.8)T^2 = 4.9T^2\) metres in \(T\) seconds. The sound then travels the same distance up the well in the remaining \(2.5 - T\) seconds, covering \(340(2.5 - T)\) metres. So \(4.9T^2 = 340(2.5 - T)\).
(c)Answer: 28.6 m
M1 for solving \(4.9T^2 + 340T - 850 = 0\)
A1 for \(T = 2.416\) (rejecting the negative root)
A1 for 28.6 (m) (accept 29)
Worked solution: \(4.9T^2 + 340T - 850 = 0 \Rightarrow T = \dfrac{-340 + \sqrt{340^2 + 4(4.9)(850)}}{9.8} = 2.416\) (the other root is negative). Depth \(= 4.9 \times 2.416^2 = 28.6\) m.
(d)Answer: Smaller: air resistance slows the stone, so in the same time it falls a shorter distance.
B1 for smaller
B1 for a reason: air resistance reduces the stone's acceleration and speed, so it falls less far in the time available
Worked solution: Smaller. With air resistance the stone accelerates less, so it is moving more slowly and falls a shorter distance in the same time.