Fig. 1 shows a velocity-time graph from a data logger fitted to an electric delivery van. The van travels along a straight warehouse road. From 0 s to 5 s i...
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Fig. 1 shows a velocity-time graph from a data logger fitted to an electric delivery van. The van travels along a straight warehouse road. From 0 s to 5 s its velocity rises steadily from zero to 4.0 m/s. It then travels at constant velocity until 12 s. The driver then releases the accelerator and the van slows. The graph is used to compare acceleration with distance travelled.
(a) What is the acceleration during the first 5.0 s? [2]
(b) Which interval has zero acceleration? [1]
(c) Sketch a horizontal line on the graph to show a van that remains stationary from 0 s to 5 s. [1]
Fig. 1 shows a train approaching a buffer stop at the end of a platform. The driver uses the brakes so the train slows uniformly from 8.0 m/s to rest in 20 s. A passenger standing in the train feels a force towards the front during the braking. The train has a mass of 1.6 x 105 kg.
(a) Calculate the acceleration of the train. [2]
(b) What is the direction of the resultant force on the train while it slows? [1]
(c) Which property of the passenger causes the passenger to move forwards relative to the train? [1]
Answer Details
Delivery van
(a) \(a=\Delta v/\Delta t=4.0/5.0=0.80\text{ m/s}^2\). [2]
(b) From 5 s to 12 s the velocity is constant, so acceleration is zero. [1]
(c) A stationary van has velocity \(0\), so the line is horizontal on the time axis from 0 s to 5 s.
[1]
Train braking
(a) \(a=(v-u)/t=(0-8.0)/20=-0.40\text{ m/s}^2\). The negative sign means acceleration is opposite to the original motion. [2]
(b) The resultant force is backwards, opposite to the train's motion. [1]
(c) The passenger moves forwards relative to the train because of inertia. [1]
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