# Finding acceleration of ball

1. Sep 15, 2004

### bob4000

how would i find acceleration if a squash ball is travelling at 9m/sec horizontally to the right and is stopped by a racket in 0.003 secs?

2. Sep 15, 2004

$$a = \frac{V_{final} - V_{initial}}{t}$$

sub in your variables and voila.

3. Sep 15, 2004

### bob4000

i still dont get it, im new to this stuff, ur gonna have to tell me step by step... and also anotha question - a tennis ball moving to the right at 5m/sec is hit by a racket and accelerated to the left leaving the tennis racket at a speed of 25m/sec to the left. if the contact time is 0.012 secs, calculate the average acceleration - again i need step by step instructions to understand

4. Sep 15, 2004

### Chi Meson

Acceleration is the rate of change of velocity, or, the total change in velocity over the time it takes to make that change.

As a ball approaches a racket, it goes from a positive velocity to a stop. Then as it rebounds it goes from zero to a negative velocity. THe total change in velocity is the amound of positive velocity it lost, plus the subsequent negative velocity it gained. In formula form it is v(final) - v (initial). If initial is a positive value, and final is a negative value, notice that your total change in velocity will be negative.

Take the total change in velocity and divide it by the time given, and you will have your acceleration. If you do it right , you will see that the acceleration is a negative quantity.

5. Sep 15, 2004

### bob4000

thankx for the help, but i think i got it wrong, cos i got -1666.6666... Should I be doin (5-25) divided by 0.012?

6. Sep 15, 2004

### Staff: Mentor

Follow exactly what Chi Meson explained....

Think this way:
initial velocity = +5 m/s (let positive mean to the right)
final velocity = -25 m/s (to the left)
change in velocity = final - initial = -25 - 5 = - 30 m/s
time = 0.012 s
average acceleration = - 30/0.012 m/s/s (note that acceleration is negative, meaning: to the left)

7. Sep 15, 2004

### bob4000

thanks to everyone for the help, really appreciate it.