Question

Honda Activa accelerates uniformly from rest to a speed of 10m/s over a distance of 25m. Determine the acceleration of the bike

Answer 1

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Abhaygupta12345

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let acceleration of the bike is a m/sec^2

bike started from rest so its initial speed will zero i.e.u=0

It reaches  at a speed of 7.10 m/s  i.e final speed  v= 7.10 m/s

In reaching speed 7.10 m/s ,it covered distance 35.4 m.

Apply equation of motion which relates u ,v ,a and s

`v^2=u^2+2as`

`(7.10)^2=(0)^2+2 a (35.4)`

`50.41=70.8a`

`a=0.712ms^(-2)`

So acceleration of the bike which started from rest , reaches at speed of 7.10 m/s by travelling distance 35.4 m will be 0.712 m/sec^2.

I hope it will help u

Answer 2

Answer:

Initial velocity = 0m/s

Speed = 10m/s

Distance = 25m

Time = distance/speed

Time = 25/10

Time = 2.5s

Acceleration = v-u/t

Acceleration = 10-0/2.5

Acceleration = 10/2.5

Acceleration = 4m/s²

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