Question

A car moving at a speed of 50km/hr can be stopped by brakes after at least 6M. If the same car is moving at a speed of 100km/hr what is the minimum stopping distance?​

Answer 1

Explanation:

Initial velocity, u = 50 km/h = 50 × (5/18) = 250/18 m/s

Final velocity, v = 0

Distance travelled before coming to rest, s = 6 m

Using, v2 = u2 + 2as

=> a = -u2/(2s)

=> a = -16.075 m/s2

Again,

u = 100 km/h = 500/18 m/s

v = 0

a = -16.075 m/s2

Now, v2 = u2 + 2as

=> s = -u2/(2a) = 24 m

Answer 2

Answer:

24m

Explanation:

The acceleration (a) is constant while the brake. so,

let u = 50 km/h and w = 100 km/h,

s = 6m and x =....

${u}^{2} = 2as$

as the v is 0

and

${w}^{2} = 2ax$

we can divide the equations...and then....

$x = { (\frac{w}{u} )}^{2}(s)$

and then placing the numbers...

$x = {(2)}^{2} (6)m$

$x = 24m$

So, the minimum stopping distance should be 24.