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June 2020 p42 q1
3809
A tram starts from rest and moves with uniform acceleration for 20 s. The tram then travels at a constant speed, \(V \text{ m s}^{-1}\), for 170 s before being brought to rest with a uniform deceleration of magnitude twice that of the acceleration. The total distance travelled by the tram is 2.775 km.
(a) Sketch a velocity-time graph for the motion, stating the total time for which the tram is moving. [2]
(b) Find \(V\). [2]
(c) Find the magnitude of the acceleration. [2]
Solution
(a) The velocity-time graph is a trapezium with the deceleration steeper than the acceleration. The total time from start to stop is 200 s.
(b) The total distance travelled is given by the area under the velocity-time graph, which is a trapezium. The area is \(0.5 \times (170 + 200) \times V = 2775\). Solving for \(V\), we get:
\(0.5 \times 370 \times V = 2775\)
\(185V = 2775\)
\(V = 15 \text{ m s}^{-1}\)
(c) The acceleration \(a\) can be found using the formula \(a = \frac{V}{t}\), where \(t = 20 \text{ s}\). Thus: