Question

The front of a big wave is approaching a beach at a constant speed of 11.6ms-! When it is 30 m away from a boy on the beach, the wave starts decelerating at a constant rate of 1.6 ms-2 and the boy walks away from the sea at a constant speed of 2 ms-! Show that the wave will not reach the boy and find the minimum distance between the boy and the wave.

Answer 1

minimum distance between the boy and the wave is 1.2 m

Explanation:

Use v = u + at, with v as 0, to find how long it takes for the wave to decelerate to 0 m/s.

0 = 11.6 - 1.6t

t = 7.25s

Then find how far the wave has traveled by substituting in s = ut + 1/2 at²

s = 11.6 * 7.25 + 1/2(-1.6)(7.25²)

s = 42.05m

Find out how far the he has traveled in the same time

s = vt + 30 (since he was 30m ahead)

s = 2*7.25 + 30

s = 44.5m

The boy has traveled a greater distance so the wave does not reach by the time it decelerates to 0m/s

Find the time at which the wave reaches a speed of 2m/s.

v = u + at

2 = 11.6 - 1.6t

t = 6s

Now find the distance traveled by the wave in that time

s = ut + 1/2 at²

s = (11.6*6) + 1/2(-1.6)(6²)

s = 40.8m

Now find the distance traveled by the boy in that same time

s = vt + 30

s = 2*6 + 30

s = 42m

Find the difference, which is the minimum separation

42 - 40.8 = 1.2m