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Question

A boy sledding down a hill accelerates at 1.20 m/s2. If he started from rest, in what distance would he reach a speed of 9.50 m/s?

please show equation used and how you got the answer

Answer 1

Firstly, we must find the time taken to reach such a velocity.

Consider the equation t = (v - v0)/a

where t = time, v = final velocity, v0 = initial velocity, a = acceleration

but v0 = 0m/s as the boy started from rest

so our equation becomes t = v/a

t = (9.5m/s)/1.2m/s/s

t = 7.9 s

Now substituting this value into the equation s = 1/2(at^2)

where s = displacement

we obtain s = 1/2(1.2m/s/s x 7.9s^2)

s = 37.4m

And thus, the problem is solved

Answer 2

Since there is no initial velocity, velocity is simply acceleration times time

9.5 = 1.2*T
T = 9.5/1.2 = 7.9166666...

Now that you know the time, you can compute position
P = 1/2a*T^2

P = 1/2*1.2*7.19666^2 = 37.6

Answer 3

4.7 meters.

I guessed.