Linear motion calculation 2
A bike has to move along a straight road. It starts from rest and then moves with uniform acceleration. After moving a distance of 100 m, it achieves a velocity of 120 m s^-1.
Find:
a) the bike acceleration
b) the time taken
c) the bike's velocity when t= 3s
SOLUTION
For question A)
Extract all the information that are needed.
u = 0 m/s
v = 120 m/s
s =100 m
What you need to find is the acceleration.
However there are several linear motion equation that needs to be considered.
1. a = (v -u)/t
2. s = 1/2 (u +v) t
3. s = ut + 1/2(at^2)
4. v^2 = u^2 + 2as
So we choose this equation v^2 = u^2 + 2as, because we have v, u and s.
rearrange the equation, we will get
a = (v^2 - u^2)/2s
= (120^2 - 0^2) / 2 (100)
= 72 m/s^2
For question B)
using a = (v - u) / t
rearrange the equation and you will get
t = (v -u)/ a
= (120 -0) / 72
= 5/3 s
For question C)
we need to find v,
so the equation that is most suitable is
v = u + at
= 0 + 72 (3)
= 216 m/s
Hope everything makes sense!
8 comments:
Why did 120 cm s^-1 become 120 m/s?
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Hi... yeah, thanks for the correction, I have already removed the cm to m. so now it's 120 m/s. again, Thanks
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