2013-12-09

Linear Motion Calculation 2

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!





6 comments:

Amer Delilbasic said...

Why did 120 cm s^-1 become 120 m/s?

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O'Deen said...

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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