Question

A bike starting from rest with a velocity of 72 km per hour over a distance of 40 metres. Calculate is acceleration. ​

Answer 1

Answer:u=0

v=72km/ph or 72*5/18 =20m/s

d=40 metres

So by third equation of motion we get

2as=v^2+u^2

2*a*40=(20)^2 + 0^2

80a=400+0

80a=400

a=400/80

a=5m/s^2

Explanation:

Answer 2

Answer:

A bike starts from rest with a velocity of 72 km per hour over a distance of 40 meters. Its acceleration is $5 \ m/s^{2}$

Explanation:

Given,

Initial velocity (u)   =  0 m/s

Final velocity (v)    =  72 km/hr  = $72\times\frac{5}{18} = 20\ m/s$

Distance covered by the bike (S) = 40 m

We have the third equation of motion as,

$v^{2}=u^{2}+2 a s$

Rearrange the above equation,

$a = \frac{v^{2} - u^{2} }{2S}$

   = $\frac{20^{2} -0 ^{2} }{2\times40} = 5 \ m/s^{2}$

Ans :

Acceleration = $5\ m/s^{2}$