Pls help me with this problem
1 answer:
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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
