SATHEE: Motion in a Straight Line

Motion in a Straight Line - Result Question 13

13. The motion of a particle along a straight line is described by equation :

$x = 8 + 12 t - t^{3}$

where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero, is :

[2012]

(a) $24 m s^{- 2}$

(b) zero

(d) $12 m s^{- 2}$

Show Answer

Answer:

Correct Answer: 13. (d)

Solution:

(d) $x = 8 + 12 t - t^{3}$

The final velocity of the particle will be zero, because it retarded.

$∴ v = 0 + 12 - 3 t^{2} = 0$

$3 t^{2} = 12$

$t = 2 s e c$

Now the retardation

$a = \frac{d v}{d t} = 0 - 6 t$

$a [t = 2] = - 12 m / s^{2}$

retardation $= 12 m / s^{2}$

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Motion in a Straight Line - Result Question 13

13. The motion of a particle along a straight line is described by equation :

Answer:

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