Question

A motorbike is traveling to the left with a speed of 27.0 m s 27.0 s m ​ 27, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction when the rider slams on the brakes. The bike skids 41.5 m 41.5m41, point, 5, start text, m, end text with constant acceleration before it comes to a stop. What was the acceleration of the motorbike as it came to a stop?

Answer 1

The acceleration of the motorbike as it came to a stop was approximately -8.77 m/s^2, indicating it was decelerating.

Step-by-step explanation:

To find the acceleration of the motorbike as it comes to a stop, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance covered:

v2 = u2 + 2as

Where:

• v is the final velocity (0 m/s since the bike stops),
• u is the initial velocity (27.0 m/s),
• a is the acceleration,
• s is the distance covered (41.5 m).

After rearranging the equation to solve for acceleration a, we get:

a = (v2 - u2) / (2s)

Substituting our known values:

a = (0 - (27.0)^2) / (2 * 41.5)

a = (-729) / (83)

a ≈ -8.77 m/s2

The negative sign indicates that the acceleration is in the opposite direction to the motion (deceleration).