A bicyclist on an old bike (combined mass: 92 kg) is rolling down (no pedaling or braking) a hill of height 120 m. Over the course of the 384 meters of downhill road, she encounters a constant friction force of 261 Newton. If her speed at the top of the hill is 9 m/s, what is her speed at the bottom of the hill

V = 48.49m/s

Explanation:

Given the following information:

Combined mass = 92kg

Hill's height = 120m

Course of ride = 384m

Frictional force = 261N

Initial speed (u) = 9m/s

Final speed (v) = ?

Since we are looking for her speed at the bottom,

Time = distance/speed = 384m/9m.s

Time = 42.67s

we use the equation

H = V²/2g ( equation for maximum heigh of trajectory)

Therefore, plugging the values we have

120 = V²/2×9.8

V² = 9.8×120×2

V = √2352

V = 48.49m/s

The acceleration of the car was one meter per second squared.

Explanation:

We can use one of the kinematic equations:

Where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time (in seconds).

The final velocity is 40 m/s, the initial velocity is 35 m/s, and the time is 5 s. Substitute and solve for a:

In conclusion, the acceleration of the car was one meter per second squared.