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 ?
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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
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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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