Question

A ball is launched from a spring on the floor with an initial velocity of 32 m/s. It then comes to a stop in 4.3 seconds. Determine the acceleration of the ball.

Answer 1

The acceleration of the ball is
`\( -7.44 \ \mathrm{m/s^2} \).`

Step-by-step explanation:

The acceleration of the ball can be determined using the kinematic equation
`\(v = u + at\), where \(v\)` is the final velocity,
`\(u\)` is the initial velocity,
`\(a\)` is the acceleration, and
`\(t\)` is the time. In this scenario, the ball comes to a stop, so the final velocity
`(\(v\))` is 0 m/s, the initial velocity
`(\(u\))` is 32 m/s (given), and the time
`(\(t\))` is 4.3 seconds (given).

Now, substituting these values into the kinematic equation, we get
`\(0 = 32 + a * 4.3\). Solving for \(a\), we find \(a = (-32)/(4.3) = -7.44 \ \mathrm{m/s^2}\).` The negative sign indicates that the acceleration is directed opposite to the initial velocity, meaning the ball is decelerating.

This negative acceleration can be interpreted as the result of forces acting in the opposite direction to the initial motion. For instance, air resistance or friction might be slowing down the ball. The magnitude of the acceleration, 7.44
`\ \mathrm{m/s^2}`, indicates the rate at which the ball's velocity decreases per unit of time. In this case, the ball is slowing down by 7.44 meters per second every second until it comes to a complete stop.