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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.

Sagot :

Winkdudeid
Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²
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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

[tex]{ \rm \sf \pink{(7.10)^2=(0)^2+2 a (35.4)}}[/tex]

[tex]{ \rm \sf \green{50.41-70.8a}}[/tex]

[tex]{ \rm \sf \pink{a=0.712ms^(-2)}}[/tex]

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.

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