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Standard XII

Physics

Order of Magnitude

A body moving...

Question

A body moving along the positive $x - a x i s$ with uniform acceleration of $- 4 m s^{- 2}$ . Its velocity at $x = 0$ is $10 m s^{- 1}$ . The time taken by the body to reach a point at $x = 12 c m$ is:

(2 s)

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(0.25 s)

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(3 s)

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(0.0125 s)

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Solution

The correct option is D $(0.0125 s)$
Acceleration $a = - 4 m / s^{2}$
Initial velocity $u = 10 m / s$
Distance traveled $x = 12 c m = 12 \times 10^{- 2} m$
From kinematics, $v^{2} - u^{2} = 2 a x$
$v^{2} = u^{2} + 2 a x$
$= 100 + 2 \times (- 4) \times 12 \times 10^{- 2}$
$= 100 - 0.96 = 99.04$
$v = 9.95 m / s$
$v = u + a t$
$t = \frac{v - u}{a}$
$= \frac{9.95 - 10}{- 4} = 0.0125 s$

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A body moving along the positive x−axis with uniform acceleration of −4ms−2. Its velocity at x=0 is 10ms−1. The time taken by the body to reach a point at x=12cm is:

The correct option is D (0.0125s)Acceleration a=−4m/s2Initial velocity u=10m/sDistance traveled x=12cm=12×10−2mFrom kinematics, v2−u2=2axv2=u2+2ax=100+2×(−4)×12×10−2=100−0.96=99.04v=9.95m/sv=u+att=v−ua=9.95−10−4=0.0125s

A body moving along the positive $x - a x i s$ with uniform acceleration of $- 4 m s^{- 2}$ . Its velocity at $x = 0$ is $10 m s^{- 1}$ . The time taken by the body to reach a point at $x = 12 c m$ is: