Question

the stopping distance of a car increases by 21%.then the incresing in speed of the car is?​

Answer 1

Answer:

$stopping \: distance \: ( s \: ) \: = \frac{u {}^{2} }{2a}$

's' is directly proportional to u^2

Distance is increased by 21%, S = 21÷100 = 0.21

u^2 = 0.21×0.21

= 0.0441

= 0.0441 ÷ 100

= 4.41 %

So, speed increases by 4.41 % when distance is increased by 21%.

I hope this helps you :)