asked 75.1k views
0 votes
24-A ball starts from rest and moves horizontally. Five positions of the ball are shown at time intervals

of 0.50 ms. The horizontal distance between X, the initial position, and Y, the final position, is 0.060 m.

What is the average acceleration of the ball between X and Y?

1 Answer

6 votes

Answer:

To calculate the average acceleration of the ball between points X and Y, you can use the following formula:

\[a_{\text{avg}} = \frac{{\Delta v}}{{\Delta t}}\]

Where:

- \(a_{\text{avg}}\) is the average acceleration.

- \(\Delta v\) is the change in velocity.

- \(\Delta t\) is the change in time.

Since the ball starts from rest, its initial velocity (\(v_i\)) is 0 m/s. We need to find the final velocity (\(v_f\)) and the time interval (\(\Delta t\)) between X and Y.

Given that the horizontal distance between X and Y is 0.060 m and we have positions at time intervals of 0.50 ms (which is 0.0005 seconds), we can calculate the final velocity using the formula:

\[v_f = \frac{{\Delta x}}{{\Delta t}}\]

Where:

- \(\Delta x\) is the change in position, which is 0.060 m.

- \(\Delta t\) is the time interval, which is 0.0005 seconds.

Now, calculate \(v_f\):

\[v_f = \frac{{0.060\, \text{m}}}{{0.0005\, \text{s}}} = 120\, \text{m/s}\]

Now that we have \(v_i = 0\, \text{m/s}\) and \(v_f = 120\, \text{m/s}\), you can calculate the average acceleration:

\[a_{\text{avg}} = \frac{{v_f - v_i}}{{\Delta t}} = \frac{{120\, \text{m/s} - 0\, \text{m/s}}}{{0.0005\, \text{s}}} = 240,000\, \text{m/s}^2\]

So, the average acceleration of the ball between X and Y is \(240,000\, \text{m/s}^2\).

answered
User KingRadical
by
7.3k points

No related questions found