Question

show that if the velocity of body increase to n times then the stopping distance increase to n2 times for uniform acceleration ​

Answer 1

Given:

Acceleration is uniform.

Velocity increases by n times.

To find:

Factor by which stopping distance increases

Step-by-step explanation:

Equation of motion is: $v^{2} = u^{2} + 2as$

u = 0

Now,

Initially,  $v^{2} = 2as$  ...........(1)

Finale, when velocity increases n times.

           $(nv)^{2} = 2as'$  .............(2)

Acceleration will be the same in 1 and 2 because acceleration is uniform.

Dividing equation 1 and 2 we get,

$s' = n^{2} s$

Hence, proved that when the velocity of a body in uniform motion increases by $n$ time, then the stopping distance increases by $n^{2}$ times.

Answer 2

Answer:

answer will be text the photos

Step-by-step explanation:

Mark me as brainlist plzz follow me e