Question

A sedan car of mass 200kg is moving with a certain velocity . It is brought to rest by the

application of brakes, within a distance of 20m when the average resistance being offered

to it is 500N.What was the velocity of the motor car?​

Answer 1

Explanation:

F=ma

500=200*a

a=2.5

now

v^2=u^2 +2as

0=u^2-2*2.5*20

u^2=100

u=10m/s

Answer 2

$\sf\star\bold\red{\underline{{QUESTION:-}}}$

✓✓A sedan car of mass 200kg is moving with a certain velocity . It is brought to rest by the application of brakes, within a distance of 20m when the average resistance being offered to it is 500N.What was the velocity of the motor car?

$\sf\star\bold\orange{\underline{{GIVEN:-}}}$

• $\sf\:Mass\:of\:car\:(m)\:=200\:kg$

• $\sf\:final\: velocity\:(v)\:=0m/s$

• $\sf\: Distance\: travelled\:(s)\:=20m$

$\sf\star\bold\green{\underline{{SOLUTION:-}}}$

$\sf\therefore\: Acceleration\:(a)\:=\dfrac{f}{m}$

$\sf\implies\:a=\dfrac{-500}{200}$

$\sf\purple{{\implies\:a=-2.5m/{s}^{2}}}$

$\sf\boxed\star\pink{\underline{\underline{{Using\:third\: equation\:of\: motion:-}}}}$

$\sf\:{v}^{2}-{u}^{2}=2as$

$\sf\implies\:{0}^{2}-{u}^{2}=2\times\:(-2.5)\times\:20$

$\sf\implies\:100={u}^{2}$

$\sf\implies\therefore{u}^{2}=100$

$\sf\implies\:u={ \sqrt{100}}$

$\sf\blue{{\implies\:u=10m/s}}$