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Standard IX

Physics

Acceleration

An object wit...

Question

An object with a mass $10 k g$ moves at a constant velocity of $10 m / s$ . A constant force then acts for $4 s e c o n d$ on the object and gives it a speed of $2 m / s$ in opposite direction. The acceleration produced is:

3 m / s e c^{2}

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- 3 m / s e c^{2}

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0.3 m / s e c^{2}

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- 0.3 m / s e c^{2}

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Solution

The correct option is B $- 3 m / s e c^{2}$
Initial velocity of the object $V_{1} = 10 m / s$
Final velocity of object $V_{2} = - 2 m / s$
Time for which the force acts $t = 4 s$
Thus acceleration of the object $a = \frac{V_{2} - V_{1}}{t} = \frac{- 2 - 10}{4} = - 3 m / s^{2}$

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An object with a mass 10kg moves at a constant velocity of 10m/s. A constant force then acts for 4second on the object and gives it a speed of 2m/s in opposite direction. The acceleration produced is:

The correct option is B −3m/sec2Initial velocity of the object V1=10m/sFinal velocity of object V2=−2m/sTime for which the force acts t=4 sThus acceleration of the object a=V2−V1t=−2−104=−3m/s2

An object with a mass $10 k g$ moves at a constant velocity of $10 m / s$ . A constant force then acts for $4 s e c o n d$ on the object and gives it a speed of $2 m / s$ in opposite direction. The acceleration produced is:

The correct option is B $- 3 m / s e c^{2}$
Initial velocity of the object $V_{1} = 10 m / s$
Final velocity of object $V_{2} = - 2 m / s$
Time for which the force acts $t = 4 s$
Thus acceleration of the object $a = \frac{V_{2} - V_{1}}{t} = \frac{- 2 - 10}{4} = - 3 m / s^{2}$