Question

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

Answer 1

V=at
S=at²/2
S=Vt/2
t=2S/V
t=35.4*2/7.10=9.97 s
a=V/t
a=7.10/9.97=0.712 m/s²
Acceleration of the bike is 0.712 m/s²

Answer 2

Explanation:

GiVeN

• A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m. Determine the acceleration of the bike

T0 FInD

• Acceleration

SoLuTi0N

• 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+2 as

• ${ \rm \sf \pink{(7.10)^2=(0)^2+2 a (35.4)}}$

• ${ \rm \sf \green{50.41-70.8a}}$
• ${ \rm \sf \pink{a=0.712ms^(-2)}}$

FiNaL AnsWeR

• 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.