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:

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

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The article is good to read thanks for sharing with us Secondary Schools in Hyderabad

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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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