Math, asked by aditim3233, 11 months ago

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

Answered by nikitavasireddi
4

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

Answered by amitnrw
0

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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