O-LevelPhysicsForces5054/22 · Oct/Nov 2016 · Q17 Marks

A skier sets off from rest and accelerates uniformly at 3.4m/s² in a straight line for 5.0s.

(a)Calculate the speed of the skier after 5.0 s. speed = ..............................................................[2][2]
(b)At 5.0s, the skier stops accelerating and travels on for a further 10.0s at a constant speed.
(b)(i)State the size of the resultant force acting on the skier during these 10.0s. ...............................................................................................................................................[1][1]
(b)(ii)On Fig. 1.1, sketch a speed-time graph for the skier during the whole 15.0s. [Figure 1.1][3]
(b)(iii)State how the distance travelled by the skier can be determined using the speed-time graph. ............................................................................................................................................... ...............................................................................................................................................[1][1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

[C1] (v = u + )at or 3.4 × 5.0
[A1] 17m/s
[B1] 0 or zero or no resultant force
[B1] straight line of positive gradient from (0, 0)
[B1] horizontal line at v > 0 and after initial acceleration
[B1] straight line from (0, 0) to (5.0, 17) and straight line from (5.0, 17) to at least (15.0, 17)
[B1] calculate the area under the graph or area of trapezium

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