The car, moving uniformly accelerated, 5 seconds after the start of the movement

The car, moving uniformly accelerated, 5 seconds after the start of the movement, reached a speed of 36 km / h. What way did the car go in 3 seconds of movement?

t = 5 s.

V0 = 0 m / s.

V = 36 km / h = 10 m / s.

t1 = 3 s.

t2 = 4 s.

S3 -?

S3 = S2 – S1, where S2 is the vehicle’s path in 4 seconds, S1 is the vehicle’s path in 3 seconds.

The path S1 with uniformly accelerated movement is determined by the formula: S1 = V0 * t1 + a * t1 ^ 2/2, where V0 is the initial speed of movement, t1 is the time of uniformly accelerated movement, and is the acceleration of the car during acceleration.

The acceleration of the body a is expressed by the formula: a = (V – V0) / t.

Since the car starts its motion from a state of rest V0 = 0 m / s, the formulas will take the form: S1 = a * t1 ^ 2/2, a = V / t.

S1 = V * t1 ^ 2/2 * t.

S1 = 10 m / s * (3 s) ^ 2/2 * 5 s = 9 m.

S2 = V * t2 ^ 2/2 * t.

S2 = 10 m / s * (4 s) ^ 2/2 * 5 s = 16 m.

S3 = S2 – S1 = 16 m – 9 m = 7 m.

Answer: in the third second of movement, the car covered the path S3 = 7 m.