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Motion in 1 dimension

  • Thread starter SarahV
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:confused: :confused: 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?
 

Answers and Replies

94
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aceeleration=change in velocity/change in time so:
17-8/3=a
d=at
so d=a * 3 seconds
 
62
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SarahV said:
:confused: :confused: 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 [tex] v^2 - u^2 = 2as [/tex]. u is the initial velocity, in this case 8 m/s. We know [tex] a = 3 m/s^2 [/tex], therefore [tex] 17^2 - 8^2 = 2(3)s [/tex] and so [tex] s = (17^2 - 8^2)/6 [/tex] which turns out to be 37.5 m.
 
33
1
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.
 
Last edited:

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