The cyclist began to climb the hill at a speed of 36 km / h, and finished at a speed of 18 km / h.

The cyclist began to climb the hill at a speed of 36 km / h, and finished at a speed of 18 km / h. How long did the ascent take and what was the cyclist’s acceleration module if he climbed 75 m?

V0 = 36 km / h = 10 m / s.

V = 18 km / h = 5 m / s.

S = 75 m.

t -?

a -?

For uniformly accelerated braking, the cyclist’s acceleration a is expressed by the formula: a = (V0 ^ 2 – V ^ 2) / 2 * S, where V0, V is the cyclist’s speed at the beginning and end of the ascent, S is the ascent path.

a = ((10 m / s) ^ 2 – (5 m / s) ^ 2) / 2 * 75 m = 0.5 m / s2.

According to the definition, acceleration a during deceleration is determined by the formula: a = (V0 – V) / t, where t is the time of speed change or the time of movement along the rise.

t = (V0 – V) / a.

t = (10 m / s – 5 m / s) / 0.5 m / s2 = 10 s.

Answer: the ascent lasted time t = 10 s, the cyclist’s acceleration a = 0.5 m / s2.