A car moving with a speed of 50 km h- 1 can be stopped by brakes in at least 6 m . If the same car is moving at a speed of 100 km h- 1 , then the minimum stopping distance (in m ) is

Question

A car moving with a speed of 50 km h- 1 can be stopped by brakes in at least 6 m . If the same car is moving at a speed of 100 km h- 1 , then the minimum stopping distance (in m ) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The stopping distance of a car is s=(u2/2 a) Therefore, for the two situations given in the question, we can write (s1/s2)=((u1/u2))2=(1/4) (6/s2)=(1/4)⇒ s2=24 m

Q. A car moving with a speed of $50 km h^{- 1}$ can be stopped by brakes in at least $6 m$ . If the same car is moving at a speed of $100 km h^{- 1}$ , then the minimum stopping distance (in $m$ ) is

The stopping distance of a car is
$s = \frac{u ^{2}}{2 a}$
Therefore, for the two situations given in the question, we can write
$\frac{s _{1}}{s _{2}} = (\frac{u _{1}}{u _{2}})^{2} = \frac{1}{4}$
$\frac{6}{s _{2}} = \frac{1}{4} \Rightarrow s_{2} = 24 m$

Q. A car moving with a speed of 50kmh−1 can be stopped by brakes in at least 6m . If the same car is moving at a speed of 100kmh−1 , then the minimum stopping distance (in m ) is

The stopping distance of a car is s=2au2​ Therefore, for the two situations given in the question, we can write s2​s1​​=(u2​u1​​)2=41​ s2​6​=41​⇒s2​=24m

Q. A car moving with a speed of $50 km h^{- 1}$ can be stopped by brakes in at least $6 m$ . If the same car is moving at a speed of $100 km h^{- 1}$ , then the minimum stopping distance (in $m$ ) is

The stopping distance of a car is
$s = \frac{u ^{2}}{2 a}$
Therefore, for the two situations given in the question, we can write
$\frac{s _{1}}{s _{2}} = (\frac{u _{1}}{u _{2}})^{2} = \frac{1}{4}$
$\frac{6}{s _{2}} = \frac{1}{4} \Rightarrow s_{2} = 24 m$