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Standard IX

Physics

Derivation of Velocity-Time Relation by Graphical Method

A ball is ini...

Question

A ball is initially moving with a velocity $0.5 m s^{- 1}$ . Its velocity decreases at a rate of $0.05 m s^{- 2}$ . How much time will it take to stop?

5 s

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10 s

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15 s

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20 s

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Solution

The correct option is B $10 s$
Given,
$u = 0.5 m s^{- 1}$
$a = - 0.05 m s^{- 2}$ and $v = 0$

Now, $v = u + a t$
$0 = 0.5 + (- 0.05 \times t)$
$t = 10 s$

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A ball is initially moving with a velocity 0.5ms−1. Its velocity decreases at a rate of 0.05ms−2. How much time will it take to stop?

The correct option is B 10sGiven,u=0.5 ms−1a=−0.05 ms−2 and v=0Now, v=u+at0=0.5+(−0.05×t)t=10 s

A ball is initially moving with a velocity $0.5 m s^{- 1}$ . Its velocity decreases at a rate of $0.05 m s^{- 2}$ . How much time will it take to stop?

The correct option is B $10 s$
Given,
$u = 0.5 m s^{- 1}$
$a = - 0.05 m s^{- 2}$ and $v = 0$

Now, $v = u + a t$
$0 = 0.5 + (- 0.05 \times t)$
$t = 10 s$