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

Question

A car moving with a speed of 40 km/h be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance: (A) 6m (B) 4 m (C) 2m (D) 8 m

Tardigrade Admin · Accepted Answer

upsilon 2=u2-2as (from equation of motion) s=(u2/2a)=(mu2/2am)=(mu2/2F) s propto u2 Hence (s1/s2)=( (u1/u2) )2 or (2/s2)=(1/4) s2=8 m

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

6m

4 m

2m

8 m

$υ^{2} = u^{2} - 2 a s$ (from equation of motion) $s = \frac{u ^{2}}{2 a} = \frac{m u ^{2}}{2 am} = \frac{m u ^{2}}{2 F}$ $s \propto u^{2}$ Hence $\frac{s _{1}}{s _{2}} = (\frac{u _{1}}{u _{2}})^{2}$ or $\frac{2}{s _{2}} = \frac{1}{4}$ $s_{2} = 8 m$

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

6m

4 m

2m

8 m

υ2=u2−2as (from equation of motion) s=2au2​=2ammu2​=2Fmu2​ s∝u2 Hence s2​s1​​=(u2​u1​​)2 or s2​2​=41​ s2​=8m

$υ^{2} = u^{2} - 2 a s$ (from equation of motion) $s = \frac{u ^{2}}{2 a} = \frac{m u ^{2}}{2 am} = \frac{m u ^{2}}{2 F}$ $s \propto u^{2}$ Hence $\frac{s _{1}}{s _{2}} = (\frac{u _{1}}{u _{2}})^{2}$ or $\frac{2}{s _{2}} = \frac{1}{4}$ $s_{2} = 8 m$