Calculating Acceleration and Displacement in 1D Motion

AI Thread Summary
The motorcyclist starts with an initial velocity of 8.0 m/s and accelerates to 17.0 m/s over 3.0 seconds. The acceleration is calculated using the formula a = (v - u) / t, resulting in 3 m/s². The displacement during this time is determined using the average velocity formula, leading to a distance of 37.5 meters. Both calculations confirm the motorcyclist's movement in the original direction. The final results are an acceleration of 3 m/s² and a displacement of 37.5 m.
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?
 
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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:
: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 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 traveled is 37.5 m to his original direction.
 
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