a car moves along a straight road with constant acceleration of 5m/s^2 .Initially at 5m, it's velocity was 3m/s^1 compute it's position and velocity at t=2seconds
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s=ut+(1/2)at^2
s=3*2+(1/2)*5*2*2
=6+10=16 m.
so final position will be 16+initial position=16+5=21 m.
v=u+at=3+5*2=13m/s.
i have assumed at s=5 m,t=0 as it says initially.
done.
feel free to ask anything.
s=3*2+(1/2)*5*2*2
=6+10=16 m.
so final position will be 16+initial position=16+5=21 m.
v=u+at=3+5*2=13m/s.
i have assumed at s=5 m,t=0 as it says initially.
done.
feel free to ask anything.
helpingbuddy40:
use v^2=u^2+2as
v=0 because at maximum height value of velocity will be zero
initial velocity is 0
but what is a
so 0=u^2+2(-g)(25)
no initial velocity u have to find
final velocity is zero
a is acceleration due to gravity(g)
u^2=50g
u=sqrt(50g) answer
Answered by
2
this can be solve with the help of differentiation
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