CBSE Class 9 Answered

Let u m/s be the initial speed. To stop the car in 16 m, required retardation a is given by

 

a = u² / (2×16) = u² /32 ..............................(1)

 

After doubling the speed and maintaining the same retardation, distance S travelled before stopping the car  is given by

 

S = (2u)² /(2×a) =  2×u² / a ..............(2)

substituting for retardation a from (1) in (2),  S = 2×u² / (u² / 32) = 64 m

 

The percentage change in this distance is =[(v2 - v1)/ v1] x 100

=  [(64 - 16)/ 16] x 100 = 3 x 100 = 300 %