Answer to Question #152873 in Mechanics | Relativity for Sanulya

Question

A motar car moving with velocity 90km-1 sees a pedestrian crossing the road 100m ahead and applies the brakes. The brakes produces a retardation of 5ms-2. If the driver applies brakes 1s after he sees the pedestrian (reaction time = 1s), what is the distance to the pedestrian when the car stops?

Answer 1

Well, let's consider that

Initial velocity V= 90 km/hour= 25 meter/second.

Dacceleration a = 5 m / s²

Let's find the distance in 1 second after sewing pedestrian:

S₁=initial velocity $\times$ time= 25$\times$ 1=25m

S₂=$V\times t- \frac{a\times t \times t}{2}$ = $25\times 1- \frac{5\times 1 \times 1}{2}$ =22.5m

The distance between pedestrian and car after stopped equals to:

D= 100-S₁-S₂= 100-25-22.5=52.5 m.