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Uniformly accelerated motion
Concepts tested here
- Distance from rest
All Questions
2009 AIPMT 1 question
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A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $S_1$ and that covered in the first 20 seconds is $S_2$, thenA constant force gives a constant acceleration a; the particle starts from rest, so u = 0.
$S = ut + \frac{1}{2}at^2 = \frac{1}{2}at^2$, so $S \propto t^2$
$S_1 = \frac{1}{2}a(10)^2 = 50a$
$S_2 = \frac{1}{2}a(20)^2 = 200a$
$\frac{S_2}{S_1} = \frac{200a}{50a} = 4$
$S_2 = 4S_1$