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A ball hits a wall horizontally with a velocity of $6.0 m / s$ After hitting the wall, it rebounds horizontally with a velocity of $4.4 / s .$ If the ball remains in contact all for $0.040 \sec$ , the acceleration of the ball would be__?

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- 260

+ 260

- 26

+ 26

answer is A.

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Detailed Solution

The acceleration of the ball would be

- 260 m / s

.
Given that the Initial velocity,

u = 6.0 m / s

and the Final velocity,

v = - 4.4 m / s

(taken as negative because it is in opposite side of initial velocity)
Time,

t = 0.040

seconds.
By the first equation of motion,

v = u + at

- 4.4 = 6.0 + a (0.040)

Accleration,

a = - 260 m / s^{2}

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