Question

A car moving along a straight road with certain velocity can be brought to rest in 2s by a retarding force F1. If retarding force F2 is taken instead of F1,the car can be brought to rest in 3s. If both forces F1 & F2 act simultaneously on the same car, then the time in which car can be bought to rest is

Answer 1

Answer:

the car can be brought to rest in $\frac{6}{5}$ seconds

Explanation:

Let the mass of the car be m and its initial velocity be u

Acceleration due to force $F_1$

$a_1=\frac{F_1}{m}$

Acceleration due to force $F_2$

$a_2=\frac{F_2}{m}$

From the first equation of motion

$0=u-a_1\times 2$

$\implies a_1=\frac{u}{2}$

Similarly,

$a_2=\frac{u}{3}$

When the combined force $F_1+F_2$ is acting

acceleration

$a=\frac{F_1+F_2}{m}$

$\implies a=\frac{F_1}{m}+\frac{F_2}{m}$

$\implies a=a_1+a_2$

$\implies a=\frac{u}{2}+\frac{u}{3}$

$\implies a=\frac{5u}{6}$

Let the car stops in time t when the combined force is causing retardation

Therefore, from first equation of motion

$0=u-at$

$\implies at=u$

$\implies \frac{5u}{6}\times t=u$

$\implies t=\frac{6}{5}$ seconds

Therefore, the car will stop in $\frac{6}{5}$ seconds

Hope this helps.