Question

On a horizontal rought rood , value of coefficient of friction mu =0.4.Find the minimum time in which a distance of 400m can be covered. The car start from rest and finally comes to rest.

Answer 1

It's 20 second.

We hace given coefficient of friction mu= .4

Friction force(f)=mu into normal=mu into mg

So maximum acceleration=f/m=4

Let car accelerate &retards for time 't' with 4m/sec^2:

(1/2)at^2+(1/2)at^2=400

So t =10second

Hence minimum time will be t+t=20seconds.

Answer 2

Given:

Value of coefficient of friction $(\mu) = 0.4.$

Distance $= 400m$

To find: Minimum time in which given distance is covered.

Solution:

                  Maximum friction on horizontal rough road,

                                    $f_(max) = \mu_(mg)$

∴ Maximum acceleration or retardation of the car may be

                              $a_(max) or a= \frac{f_(max)}{m}$

                                                $= \frac{\mu mg}{m} \\ = \mu g$

                                                $=0.4\times10=4m/s^2$

Maximum acceleration or retardation of the car is  $4m/s^2$

Let , the rest car acceleration and retards for time $'t'$ with $4m/s^2$

                 Then,  

                               $\frac{1}{2} at^2+\frac{1}{2} at^2 = 400m$

                               $at^2=400m$

                              $4t^2=400$

       or                   $t=10s$

Therefore the minimum time is $20s$ ($10s$ of acceleration and $10s$ of    retardation).