The front of a big wave is approaching a beach at a constant speed of 11.6m/s. When it is 30m away from a boy on the beach, the wave starts decelerating at a constant rate of 1.6m/s² and the boy walks away from the sea at a constant speed of 2m/s. Show that the wave will not reach the boy and find the minimum distance between the boy and the wave.
Answers
Answer:
1.2m
Step-by-step 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 boy 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.
Also, the distance is minimum when the wave reaches a speed of 2m/s. Because beyond that, it gets slower than the boy and the distance between them will increase. So 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
Given : big wave is approaching a beach at a constant speed of 11.6 m/s. when it is 30 m away from a boy on the beach, the wave starts decelerating at a constant rate of 1.6 m/s^2 and the boy walks away from the sea at a constant speed of 2 m/s.
To Find : Show that the wave will not reach the boy and find the minimum distance between the boy and the wave.
Solution:
S = ut + (1/2)at²
Distance covered by the wave = 11.6t - (1/2)1.6t²
Distance covered by the boy = 2t
Distance between wave and boy = 30 - (11.6t - (1/2)1.6t² - 2t )
= 0.8t² - 9.6t +30
f(t) = 0.8t² -9.6t + 30 distance between wave any boy at any time t
f'(t) = 1.6t - 9.6
f'(t) = 0 => t = 6
f''(t) = 1.6 > 0
Hence at t = 6 distance is minimum
So minimum distance is
0.8t² - 9.6t +30
= 0.8(6)² - 9.6 * 6 + 30
= 1.2
After 6 secs the velocity of wave becomes less than boys velocity
Hence the wave will not reach the boy
the minimum distance between the boy and the wave. = 1.2 m
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