First slide

Projection Under uniform Acceleration

Question

A body sliding on a smooth inclined plane requires 4 s to reach the bottom, starting from rest at the top. How much time does it take to cover one-fourth the distance starting from rest at the top?

Moderate

A

1 s
B

2 s
C

4 s
D

16 s

Solution

When a body slides on an inclined plane, component of weight along the plane produces an acceleration.
$a = \frac{m g \sin θ}{m}$ =g sin $θ$ = Constant
If s is the length of the inclined plane, then $s = 0 + \frac{1}{2} a t^{2} = \frac{1}{2} g \sin θ . t^{2}$
$\frac{s'}{s} = \frac{t'^{2}}{t^{2}} o r \frac{s}{s'} = \frac{t^{2}}{t'^{2}}$ , Given t = 4 s and $s' = \frac{s}{4}$
$t' = t \sqrt{\frac{s'}{s}} = 4 \sqrt{\frac{s}{4 s}} = \frac{4}{2} = 2 s$

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