Question

A car is moving with a velocity of 10 m/s. The driver sees a wall ahead of him and applied brakes. The car stops after covering 10 m distance. If the car was moving with a speed pf 20 ms −1 , it would have stopped in 30 m distance. The reaction of the driver is

Answer 1

Answer:

Case I. Suppose reduction time = t                  , so = 10 meters

then distance in t sec will be 10t meter

⇒ remaining distance = 10 meter - 10 t

using v  

2

=u  

2

−2as

we get s=  

2a

u  

2

−v  

2

 

​  

=  

2a

(10)  

2

−0

​  

=  

a

50

​  

 

⇒  

a

50

​  

=10−10t _______ (1)

Case II  

u=20 m/s & S  

0

​  

=30 meter

distance in t sec will be 20 t meter

distance during retardation S=  

2a

u  

2

−v  

2

 

​  

=  

2a

(20)  

2

−0

​  

 

S=  

a

200

​  

 

S  

0

​  

=30m⇒  

a

200

​  

+20t=30

           

a

50

​  

+5t=7.5 _______ (2)

using (1)  

a

50

​  

+10t=10

           -           -         -

        ___________________

          0-5t= -2.5 ⇒ subtracting t=0.5 second

Explanation: