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June 2011 p43 q4
3717
The diagram shows the velocity-time graphs for the motion of two particles P and Q, which travel in the same direction along a straight line. P and Q both start at the same point X on the line, but Q starts to move T s later than P. Each particle moves with speed 2.5 m s-1 for the first 20 s of its motion. The speed of each particle changes instantaneously to 4 m s-1 after it has been moving for 20 s and the particle continues at this speed.
Make a rough copy of the diagram and shade the region whose area represents the displacement of P from X at the instant when Q starts. [1]
It is given that P has travelled 70 m at the instant when Q starts. Find the value of T. [2]
Find the distance between P and Q when Q's speed reaches 4 m s-1. [2]
Sketch a single diagram showing the displacement-time graphs for both P and Q, with values shown on the t-axis at which the speed of either particle changes. [2]
Solution
(i) The area representing the displacement of P from X at the instant when Q starts is a composite figure consisting of two rectangles. The first rectangle has boundaries at t = 0 and t = 20, with v = 0 and v = 2.5. The second rectangle has boundaries at t = 20 and t = T, with v = 0 and v = 4.
(ii) The displacement of P when Q starts is given by:
\(50 + 4(T - 20) = 70\)
Simplifying gives:
\(4T - 30 = 70\)
\(4T = 100\)
\(T = 25\)
(iii) The distance between P and Q when Q's speed reaches 4 m s-1 is calculated by:
\(\text{Distance} = 70 + (4 - 2.5) \times 20\)
\(\text{Distance} = 70 + 30 = 100 \text{ m}\)
(iv) For the displacement-time graph, P is represented by two straight line segments: the first with a positive slope and the second with a steeper slope, with t = 20 indicated. For Q, the first and second segments are parallel to P's and displaced to the right, with t = 25 and t = 45 indicated.