Physics, asked by Ayushsingh2387, 1 month ago

A bicycle starts from rest and moves with uniform acceleration 1.5 m/s^2 , his velocity reaches 7.5 m/s through a distance of ?

Answers

Answered by sumedhsumedh54
0

u=0

v=7.5m/s

a=1.5m/s^2

s=?

from second equation of motion

v^2 =u^2+2as

7.5^2=0 +2×1.5s

56.25= 3s

s=56.25/3 m

s=18.75 . Answer

Answered by tanyasaini2003
0

Given:

Initial velocity (u) of the bicycle= 0 m/s

Final velocity (v) of the bicycle= 7.5 m/s

Rate of uniform acceleration (a) of the bicycle= 1.5 m/s^2

To Find:

The distance through which the velocity of the bicycle reaches 7.5 m/s.

Solution:

  • By using the Third equation of motion,

             v^2 - u^2 = 2as

we can find 's' which is the distance through which the velocity of the bicycle reaches 7.5 m/s.

  • So,

⇒( 7.5)^2 - 0^2 = 2× 1.5× s

⇒56.25 = 3×s

⇒56.25÷3 = s

⇒18.75 = s

∴ s = 18.75 m

Hence, the distance through which the velocity of the bicycle reaches 7.5 m/s is 18.75 m.

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