The car started braking from the traffic light at a distance of 80 meters, having an initial speed of 90 km / h

The car started braking from the traffic light at a distance of 80 meters, having an initial speed of 90 km / h, the braking time is 4 seconds to determine the acceleration and distance to the traffic light.

Ssph = 80 m.

V0 = 90 km / h = 25 m / s.

V = 0 m / s.

t = 4 s.

a -?

Ssph1 -?

With uniformly accelerated braking, the path S traveled by the car is expressed by the formula: S = V0 * t – a * t ^ 2/2, where V0 is the speed before the start of braking, t is the braking time, a is the acceleration during braking.

Acceleration a during braking shows how the vehicle speed decreases over time t and is determined by the formula: a = (V0 – V) / t.

a = (25 m / s – 0 m / s) / 4 s = 6.25 m / s2.

S = 25 m / s * 4 s – 6.25 m / s2 * (4 s) ^ 2/2 = 50 m.

Ssph1 = Ssph – S.

Ssph1 = 80 m – 50 m = 30 m.

Answer: the car braked with acceleration a = 6.25 m / s2, stopped at a distance of Ssph1 = 30 m from the traffic light.