Question

Mechanics paper/Math (please helpp!!!)
Alan starts walking from a point O, at a constant speed of 4 m s−1, along a horizontal path. Ben walks along the same path, also starting from O. Ben starts from rest 5 s after Alan and accelerates at 1.2 m s−2 for 5 s. Ben then continues to walk at a constant speed until he is at the same point, P, as Alan.
(i) Find how far Ben has travelled when he has been walking for 5 s and find his speed at this instant. [2]
(ii) Find the distance OP. [3]

Answer 1

Answer:

  (i) 15 m, 6 m/s

  (ii) 90 m

Step-by-step explanation:

(i) For some acceleration (a) from rest, the distance covered (d) in time t is ...

  d = (1/2)at^2

The distance covered by Ben in the 5 seconds he is accelerating is ...

  d = (1/2)(1.2 m/s²)(5 s)² = 15 m

Of course, Ben's speed at that point is ...

  s = (1.2 m/s²)(5 s) = 6 m/s

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(ii) When Ben has been walking 5 s, Alan has been walking 10 s, so Alan has covered (10 s)(4 m/s) = 40 m. Their distance difference of 40 -15 = 25 m is being made up at the rate of their speed differences: (6 m/s) -(4 m/s) = 2 m/s.

It will take (25 m)/(2 m/s) = 12.5 s additional time for Ben to catch Alan. In the 22.5 s that Alan has been walking before they meet, he will have walked ...

  (22.5 s)(4 m/s) = 90 m . . . the distance OP