Question

032. A 25.0 kg bicycle is traveling at a speed of 10.0 m/s.

a. If the bike comes to rest in 2.0 s, what is the magnitude
of its acceleration?
b. What net force is needed to provide this acceleration?
c. How far will the bike travel in the 2.0 s?

Answer 1

a. The magnitude of acceleration is -5.0 m/s². b. The net force needed to provide this acceleration is -125.0 N. c. The bike will travel a distance of 10.0 m in 2.0 s.

Step-by-step explanation:

a. The magnitude of acceleration can be found using the formula: acceleration = (final velocity - initial velocity) / time. In this case, the initial velocity is 10.0 m/s and the final velocity is 0 m/s (since the bike comes to rest). The time is 2.0 s. Plugging in these values, we get: acceleration = (0 - 10.0) / 2.0 = -5.0 m/s².

b. The net force needed to provide this acceleration can be calculated using Newton's second law of motion: force = mass x acceleration. In this case, the mass is 25.0 kg and the acceleration is -5.0 m/s² (negative because it is deceleration). Plugging in these values, we get: force = 25.0 kg x -5.0 m/s² = -125.0 N (since force is a vector, negative sign indicates the opposite direction of motion).

c. The distance traveled by the bike can be calculated using the formula: distance = initial velocity x time + 0.5 x acceleration x time^2. In this case, the initial velocity is 10.0 m/s, the time is 2.0 s, and the acceleration is -5.0 m/s². Plugging in these values, we get: distance = 10.0 m/s x 2.0 s + 0.5 x -5.0 m/s² x (2.0 s)^2 = 20.0 m - 10.0 m = 10.0 m.

Answer 2

See below

Step-by-step explanation:

acceleration = change in velocity / change in time

= - 10 m/ s / 2 sec = - 5 m /s^2 = 5 m/s^2 (magnitude)

F = ma = 25 kg * 5 m/s^2 = 125 N

x = vo t + 1/2 at ^2

= 10 (2) + 1/2 ( -5)(2^2 ) = 10 meters