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Second Law of Motion

A driver firs...

Question

A driver first accelerates his car at a rate of $1.8 m / s^{2}$ and then at a rate of $1.2 m / s^{2}$ . The ratio of the forces exerted by the engines respectively will be equal to

2 : 3

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1 : 2

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2 : 1

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3 : 2

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Solution

The correct option is D 3 : 2
$First situation: - -------------- -$
Acceleration $a_{1} = 1.8 m / s^{2}$
$F_{1} = m a_{1} = m \times 1.8$

$Second situation: - ----------------- -$
Acceleration $a_{2} = 1.2 m / s^{2}$
$F_{2} = m a_{2} = m \times 1.2$

So, $\frac{F_{1}}{F_{2}} = \frac{a_{1}}{a_{2}} = \frac{1.8}{1.2} = 3 : 2$

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A driver first accelerates his car at a rate of 1.8m/s2 and then at a rate of 1.2m/s2. The ratio of the forces exerted by the engines respectively will be equal to

The correct option is D 3 : 2First situation:––––––––––––––––Acceleration a1=1.8 m/s2F1=ma1=m×1.8 Second situation:–––––––––––––––––––Acceleration a2=1.2 m/s2F2=ma2=m×1.2So, F1F2=a1a2=1.81.2=3:2

A driver first accelerates his car at a rate of $1.8 m / s^{2}$ and then at a rate of $1.2 m / s^{2}$ . The ratio of the forces exerted by the engines respectively will be equal to