A car moving with speed of $40 k m / h r$ can be stopped by applying brakes after at least $2 m$ . If the same car is moving with a speed of $80 k m / h r$ what is the minimum stopping distance?

2 m

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4 m

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6 m

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8 m

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Solution

The correct option is D $8 m$
Stopping distance $s = \frac{u^{2}}{2 a}$
$\frac{s_{2}}{s_{1}} = \frac{u_{2}^{2}}{u_{1}^{2}} \Rightarrow s_{2} = {[\frac{u_{2}}{u_{1}}]}^{2} s_{1} \Rightarrow s_{2} = {(\frac{80}{40})}^{2} 2 \Rightarrow s_{2} = 8 m$

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A car moving with a speed of $40 k m / h r$ can be stopped by applying breaks for $2$ km. If the same car is moving with a speed of $80 k m / h r$ , what is the minimum stopping distance? Assume constant retardation.

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A car moving with speed of 40km/hr can be stopped by applying brakes after at least 2m. If the same car is moving with a speed of 80km/hr what is the minimum stopping distance?

The correct option is D 8mStopping distance s=u22as2s1=u22u21⇒s2=[u2u1]2s1⇒s2=(8040)22⇒s2=8m

A car moving with speed of $40 k m / h r$ can be stopped by applying brakes after at least $2 m$ . If the same car is moving with a speed of $80 k m / h r$ what is the minimum stopping distance?

The correct option is D $8 m$
Stopping distance $s = \frac{u^{2}}{2 a}$
$\frac{s_{2}}{s_{1}} = \frac{u_{2}^{2}}{u_{1}^{2}} \Rightarrow s_{2} = {[\frac{u_{2}}{u_{1}}]}^{2} s_{1} \Rightarrow s_{2} = {(\frac{80}{40})}^{2} 2 \Rightarrow s_{2} = 8 m$