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kifflom [539]
3 years ago
15

A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

and slows at 7.0 m/s2. How far does the car go before it stops?
Physics
1 answer:
mr Goodwill [35]3 years ago
3 0

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

d_1 = v* t

d_1 = 20 * 0.8

d_1 = 16 m

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2*(-7)*d_2

d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

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Answer:

a) Knowns, initial speed v_{i}=13.0 m/s, final speed v_{f}=0 m/s and gravity due it is a constant g=9.8m/s^{2}

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c) Total time is t=2.65s

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at 0m/s speed.

b) Second, now that we know final speed we use v_{f} =v_{i}-gt, as we clear for t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s.

Then we use y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2}  =8.62m

c)Third, finally we can use y=v_{i}t-\frac{1}{2} gt^{2}, as we know y=0m when the dolphin fall into the water again and v_{i} =13.0m/s, then we have 0=(13m/s)t-\frac{1}{2} (9.8m/s^{2}  )t^{2} is a quadratic form 0=t(13.0-4.9t) so we have t_{i}=0s and t_{f}=\frac{13}{4.90}  =2.65s

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