An object moving along a straight line is initially moving at a velocity of $+ 12 m / s$ , but is accelerating so that at a time $4$ seconds later it is moving at the same speed, but in the opposite direction.
What is the object's average acceleration during the $4$ seconds?

0 m / s^{2}

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6 m / s^{2}

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- 3 m / s^{2}

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- 6 m / s^{2}

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Solution

The correct option is D $- 6 m / s^{2}$
Initial velocity of the object $v_{i} = + 12$ m/s (positive direction)
Final velocity of the object $v_{f} = - 12$ m/s (negative direction)
Thus acceleration of the object $a = \frac{v_{f} - v_{i}}{t} = \frac{- 12 - 12}{4} = - 6$ $m / s^{2}$

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An object moving along a straight line is initially moving at a velocity of +12m/s, but is accelerating so that at a time 4 seconds later it is moving at the same speed, but in the opposite direction.What is the object's average acceleration during the 4 seconds?

The correct option is D −6m/s2Initial velocity of the object vi=+12 m/s (positive direction)Final velocity of the object vf=−12 m/s (negative direction)Thus acceleration of the object a=vf−vit=−12−124=−6 m/s2

An object moving along a straight line is initially moving at a velocity of $+ 12 m / s$ , but is accelerating so that at a time $4$ seconds later it is moving at the same speed, but in the opposite direction.
What is the object's average acceleration during the $4$ seconds?

The correct option is D $- 6 m / s^{2}$
Initial velocity of the object $v_{i} = + 12$ m/s (positive direction)
Final velocity of the object $v_{f} = - 12$ m/s (negative direction)
Thus acceleration of the object $a = \frac{v_{f} - v_{i}}{t} = \frac{- 12 - 12}{4} = - 6$ $m / s^{2}$