Question

Honda activa accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. determine the acceleration of the bike

Answer 1

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

"Acceleration of the bike which started from rest, reaches at speed of 7.10 $\frac { m }{ s } by travelling distance 35.4 m will be 0.712 \frac { m }{ { sec }^{ 2 } }$.

Solution:

Let acceleration of the bike is a $\frac { m }{ { sec }^{ 2 } }$ .

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

It reaches at a speed of 7.10 $\frac { m }{ s }  i.e final speed  v = 7.10 \frac { m }{ s }$

In reaching speed $7.10 \frac { m }{ s }$, it covered distance 35.4 m.

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

${ v }^{ 2 }={ u }^{ 2 }+2as$

${ (7.10) }^{ 2 }\quad =\quad { (0) }^{ 2 }+2a(35.4)$

50.41 = 70.8a

$a\quad =\quad 0.712{ ms }^{ -2 }$"