Find acceleration from SUVAT

A question is this type if and only if it provides kinematic information and asks to find acceleration or deceleration using SUVAT equations.

5 questions · Moderate -0.8

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Edexcel M1 2002 June Q1
6 marks Moderate -0.8
A car moves with constant acceleration along a straight horizontal road. The car passes the point \(A\) with speed \(5 \text{ m s}^{-1}\) and \(4 \text{ s}\) later it passes the point \(B\), where \(AB = 50\text{m}\).
  1. Find the acceleration of the car. [3]
When the car passes the point \(C\), it has speed \(30 \text{ m s}^{-1}\).
  1. Find the distance \(AC\). [3]
Edexcel M1 Q1
7 marks Moderate -0.8
An aircraft moves along a straight horizontal runway with constant acceleration. It passes a point \(A\) on the runway with speed \(16\) m s\(^{-1}\). It then passes the point \(B\) on the runway with speed \(34\) m s\(^{-1}\). The distance from \(A\) to \(B\) is \(150\) m.
  1. Find the acceleration of the aircraft. [3]
  2. Find the time taken by the aircraft in moving from \(A\) to \(B\). [2]
  3. Find, to 3 significant figures, the speed of the aircraft when it passes the point mid-way between \(A\) and \(B\). [2]
Edexcel M1 Q4
9 marks Standard +0.3
A sports car is being driven along a straight test track. It passes the point \(O\) at time \(t = 0\) at which time it begins to decelerate uniformly. The car passes the points \(L\) and \(M\) at times \(t = 1\) and \(t = 4\) respectively. Given that \(OL\) is 54 m and \(LM\) is 90 m,
  1. find the rate of deceleration of the car. [5 marks]
The car subsequently comes to rest at \(N\).
  1. Find the distance \(MN\). [4 marks]
AQA AS Paper 1 2020 June Q13
3 marks Easy -1.8
An object is moving in a straight line, with constant acceleration \(a\text{ m s}^{-2}\), over a time period of \(t\) seconds. It has an initial velocity \(u\) and final velocity \(v\) as shown in the graph below. \includegraphics{figure_13} Use the graph to show that $$v = u + at$$ [3 marks]
AQA Paper 2 Specimen Q12
4 marks Moderate -0.8
A particle moves on a straight line with a constant acceleration, \(a\) m s\(^{-2}\). The initial velocity of the particle is \(U\) m s\(^{-1}\). After \(T\) seconds the particle has velocity \(V\) m s\(^{-1}\). This information is shown on the velocity-time graph. \includegraphics{figure_12} The displacement, \(S\) metres, of the particle from its initial position at time \(T\) seconds is given by the formula $$S = \frac{1}{2}(U + V)T$$
  1. By considering the gradient of the graph, or otherwise, write down a formula for \(a\) in terms of \(U\), \(V\) and \(T\). [1 mark]
  2. Hence show that \(V^2 = U^2 + 2aS\) [3 marks]