# Motion in 1 dimension  Help! I need to know how to work this problem!! A motorcyclist moving with an initial velocity of 8.0 m/s undergoes a constant acceleration for 3.0s, at which time his velocity is 17.0 m/s. What is the acceleration, and how far does he travel in the 3.0 s interval?

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aceeleration=change in velocity/change in time so:
17-8/3=a
d=at
so d=a * 3 seconds

SarahV said:  Help! I need to know how to work this problem!! A motorcyclist moving with an initial velocity of 8.0 m/s undergoes a constant acceleration for 3.0s, at which time his velocity is 17.0 m/s. What is the acceleration, and how far does he travel in the 3.0 s interval?
Actually, d=at is incorrect.
The correct formula is $$v^2 - u^2 = 2as$$. u is the initial velocity, in this case 8 m/s. We know $$a = 3 m/s^2$$, therefore $$17^2 - 8^2 = 2(3)s$$ and so $$s = (17^2 - 8^2)/6$$ which turns out to be 37.5 m.

8+3a=17
a=3m/s^2
8*3+0.5*3*3^2=24+13.5=37.5m

Let me summarize the solution.

a=acceleration u=initial velocity v=final velocity
s=displacement t=time taken

a=(v-u)/t
a=(17.0-8.0)/3.0
a=3 m/(s^2)

s=((u+v)*t)/2
s=((8.0+17.0)*3.0)/2
s=37.5 m

Therefore, the acceleration is 3 m/(s^2) to his original direction and
the displacement he travelled is 37.5 m to his original direction.

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