Question

A vehicle moving with a velocity of 2 m/s can be stopped over a distance of 2 m. Keeping the retarding force constant, if kinetic energy is doubled, what will be the distance covered before it comes to rest?​

Answer 1

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$\huge\mathcal\pink{Solution:-}$

⛄ Answer:-

☞ Distance covered by the vehicle before it comes to rest = 4m.

⛄ Step-by-step explanation:-

☘ Given:-

Mass = m, u= 2 m/s, s= 2m, v= 0

☘ To find:-

The distance traversed before it comes to rest.

☘ Solution:-

Using position-velocity equation of motion, we have,

v² - u² = 2as

Now, on substituting the values given, we get,

0 - 2² = 2 × a × 2

=> a = -1 m/s²

Now, K.E. of the body = 1/2 m(2)² = 2J,

Therefore,

If K.E. is doubled, then K.E. = 4

Also, if velocity of the body becomes u2, then,

K.E. = 1/2 m(u2)² = 4 J

=> u2 = 2√2

Now, let s' be the distance at which it will stop.

v² - u² = 2as'

=> 0² - (2√2)² = 2 × (-1) × s'

☞ s' = 4m.

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Answer 2

Explanation:

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