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Standard IX

Physics

Second Law of Motion

A car of mass...

Question

A car of mass $480 k g$ moving at a speed of $54 k m p h$ , is stopped in $10 s$ . Calculate the force applied by the brakes.

720

N

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700

N

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530

N

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100

N

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Solution

The correct option is A $720$ $N$
$u = 54 k m p h = \frac{54 \times 1000}{3600} m / s = 15 m / s$
$v = 0$
and, $t = 10 s$
So, $a = \frac{v - u}{t} = \frac{15 - 0}{10} = 1.5 m / s^{2}$
Mass, $M = 480 k g$
So, force: $F = 480 \times 1.5 = 720 N$

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A car of mass 480 kg moving at a speed of 54 kmph, is stopped in 10 s. Calculate the force applied by the brakes.

The correct option is A 720 Nu=54 kmph=54×10003600 m/s=15 m/sv=0and, t=10 sSo, a=v−ut=15−010=1.5 m/s2Mass, M=480 kgSo, force: F=480×1.5=720 N

A car of mass $480 k g$ moving at a speed of $54 k m p h$ , is stopped in $10 s$ . Calculate the force applied by the brakes.

The correct option is A $720$ $N$
$u = 54 k m p h = \frac{54 \times 1000}{3600} m / s = 15 m / s$
$v = 0$
and, $t = 10 s$
So, $a = \frac{v - u}{t} = \frac{15 - 0}{10} = 1.5 m / s^{2}$
Mass, $M = 480 k g$
So, force: $F = 480 \times 1.5 = 720 N$