Question

A vehicle was moving at a speed of 90 km/h. On seeing a baby 20 m ahead on the road, the driver jammed on the brakes and it came to rest at a distance of 15 m. What is its retardation and how long does it take to come at rest?
{Ans=20.83 m/s^2, 1.2 s} I want the process to do it.

Answer 1

Answer:

Initial velocity u = 90 km/h =25 m/s

Final velocity v = 0 m/s (brought to rest)

Distance travelled s in coming to rest = 25 m

Using the relation,

v² - u² = 2 a s, where a is the acceleration

0² - 25² = 2 . a . 25,

a = - ( 625/50) = - 12.5 m/s². The retardation is 12.5 m/s² ( negative sign in a implies it is retardation).

From a we can calculate time for which brakes were applied.

u = 25 m/s

v = 0 m/s

a = - 12. 5 m/s²

Using the relation, v = u + a t,

Substituting various values, we get

0 m/s =2 5 m/s - 12.5 m/s² t s

t = 2 s.

So brakes are applied for 2 s.

Answer 2

Given

⚠️ Initial speed of vehicle = 90 km/h

⚠️ Comes to rest at a distance of 15 m.

To find

⚠️ Retardation of vehicle.

⚠️ Time taken to come to rest.

Solution

As per given data we have :

⇒ Initial speed (u) = 90 km/h = (90 × 5/18) = 25 m/s

⇒ Distance travelled (s) = 15 m

⇒ Final speed (v) = 0 m/s

Now using 3rd equation of motion :

✒️ v² - u² = 2as

✒️ (0)² - (25)² = 2 × a × 15

✒️ 0 - 625 = 30a

✒️ -625 = 30a

✒️ a = -625/30

✒️ a = -20.83 m/s²

As we know, acceleration = - Retardation

Therefore, retardation = 20.83 m/s²

________________

Now using 1st equation of motion :

✒️ v = u + at

✒️ 0 = 25 + (-20.83) × t

✒️ 0 - 25 = -20.83t

✒️ -25 = -20.83t

✒️ t = -25/-20.83

✒️ t = 1.2 s

Therefore, time taken to come to rest = 1.2 seconds.