# A basketball rolls onto the court at a speed of 4.0 m/s to the right and slows down with a constant deceleration of 0.50 m/s^2 over a distance of 14 m. What is the velocity of the basketball after rolling for 14 m?

Physics · High School · Thu Feb 04 2021

Answered on

To determine the velocity of the basketball after it has rolled for a distance of 14 m with a constant negative acceleration, we can use the kinematic equation which relates initial velocity (v_i), final velocity (v_f), acceleration (a), and distance (d):

\( v_f^2 = v_i^2 + 2ad \)

Where: \( v_i \) = initial velocity = 4.0 m/s (to the right) \( a \) = acceleration = -0.50 m/s² (negative since it's slowing down) \( d \) = distance = 14 m

We are looking for the final velocity (\( v_f \)) after the basketball has traveled 14 m. Plugging in the values, we get:

\( v_f^2 = (4.0 \, m/s)^2 + 2(-0.50 \, m/s²)(14 \, m) \) \( v_f^2 = 16.0 \, m^2/s^2 - 14 \, m/s^2 \) \( v_f^2 = 16.0 \, m^2/s^2 - 14 \, m/s^2 \) \( v_f^2 = 2.0 \, m^2/s^2 \)

To find \( v_f \), we take the square root of both sides:

\( v_f \) = √(2.0 m²/s²) \( v_f \) = 1.41 m/s (approximately)

Answered on

To calculate the velocity of the basketball after rolling for 14 meters, you can use the kinematic equation that relates initial velocity, acceleration, and displacement:

\( v^2 = u^2 + 2as \)

Where: - \( v \) is the final velocity, - \( u \) is the initial velocity, - \( a \) is the acceleration, - \( s \) is the displacement.

Given: - \( u = 4.0 \, \text{m/s} \) (initial velocity to the right), - \( a = -0.50 \, \text{m/s}^2 \) (acceleration to the left, thus negative because it opposes the direction of the velocity), - \( s = 14 \, \text{m} \) (displacement).

Now, let's plug in the values:

\( v^2 = (4.0 \, \text{m/s})^2 + 2(-0.50 \, \text{m/s}^2)(14 \, \text{m}) \)

\( v^2 = 16.0 \, \text{m}^2/\text{s}^2 - 14.0 \, \text{m}^2/\text{s}^2 \)

\( v^2 = 2.0 \, \text{m}^2/\text{s}^2 \)

To find the final velocity \( v \), we take the square root of both sides:

\( v = \sqrt{2.0 \, \text{m}^2/\text{s}^2} \)

\( v = 1.41 \, \text{m/s} \)

Since we are dealing with the magnitude and the acceleration was negative (implying that it's in the opposite direction to the initial velocity), the final velocity to the right will be:

\( v = 1.41 \, \text{m/s} \, \text{(to the right)} \)