Question

A body moves over horizontal surface with initial velocity of 2m/s velocity decreases uniformly at rate of 0.25m/s^2 how much time will it take to stop

Answer 1

Answer:8sec

Explanation:v=u-at (- sign because particle is regarding) so 0=2-1/4t

1/4t=2

T=8 secs

Answer 2

Explanation:

By Formula

v=u+at

final velocity becomes 0 so

v=0

u=2 m/s

a=0.25m/s²

since the acceleration is increasing to decrease the velocity

so =(-0.25)

Now in formula

0=2+(-0.25)*t

-2=-0.25t

t=-2/(-0.25)

t=8

Thank you

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