Question 169: Laws of Motion

Question

Consider a car moving on a straight road with a speed of $$100$$ $$m/s$$. The distance at which car can be stopped is $$\left[ {{\mu _k} = 0.5} \right]$$

Accepted Answer

Acceleration due to friction = $$\left( { - {\mu _k}g} ight)$$
We know, $${v^2}  = {u^2} + 2as$$
$$ \Rightarrow $$ $${0^2} =   {u^2} + 2\left( { - {\mu _k}g} ight)s$$
$$ \Rightarrow $$ $$2 { {\mu _k}g}s$$ = $${u^2}$$
$$ \Rightarrow s = {{{{100}^2}} \over {2 	imes 0.5 	imes 10}}$$
$$ \Rightarrow s = 1000\,m$$

Answer

$$1000$$ $$m$$

Answer

$$800$$ $$m$$

Answer

$$400$$ $$m$$

Answer

$$100$$ $$m$$

Laws of Motion

Consider a car moving on a straight road with a speed of $100 m/s. The distance at which car can be stopped is \left[ {{\mu _k} = 0.5} \right]

Choose the correct answer: