A handball is hit toward a wall with a velocity of 14.3 m/s in the forward direction. It returns with a velocity of 13.5 m/s in the backward direction. If the time interval during which the ball is accelerated is 0.011 s, what is the handball's average acceleration?

Let's define t = 0 as the moment when the ball hits the wall.

in this moment, we have a given acceleration, but if we only want to calculate the average acceleration, then let's consider the acceleration constant.

a(t) = A.

To find the velocity equation we should integrate, and the constant of integration will be the initial velocity, in this case, is 14.3m/s.

v(t) = A*t + 14.3m/s

Now we know that at t = 0.011s, the velocity is -13.5m/s (the sign is negative because the ball is moving in the opposite direction as before).

Then we can solve the equation:

v(0.011s) = -13.5m/s = A*0.011s + 14.3m/s

Now we can solve it for A, the average acceleration:

-13.5m/s - 14.3m/s = A*0.011s

(-27.8m/s)/0.011s = A = -1313.5 m/s^2

real, smaller and upside-down.   so the answer should be b.