Physics, asked by sanjaykandpalpbhojd, 1 year ago

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

Answers

Answered by sonuvuce
8

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.

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