A tennis ball with a speed of 10.0m/s is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 8.0m/s. If the ball is in contact with the wall for .012s, what is the average acceleration of the ball while in contact with the wall?
2007-08-06 18:27:50 · 4 answers · asked by sara.girl122 1 in Science & Mathematics ➔ Physics
a = dv/dt; where dv = change in velocity = (10 + 8) m/sec; the two speeds are added because they are going in opposite directions. That is dv = v1 - (-v0) = v1 + v0 because v1 is going in the plus direction (to the wall) and v0 is going in the minus direction (away). dt = .012 sec, the interval of time where the velocity changed direction and speed.
Thus, a = dv/dt = 18 m/sec//.012 sec, which is the acceleration. You can do the arithmetic.
2007-08-06 18:35:26 · answer #1 · answered by oldprof 7 · 0⤊ 0⤋
Average acceleration = [Change in velocity] / time
Change in velocity = 10 â [â 8] = 10 + 8 = 18 m/s
Average acceleration = 18 / 0.012 = 1500 m/s^2
2007-08-06 18:56:55 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
It is 18 / .012 m/s^2
2007-08-06 18:41:38 · answer #3 · answered by Aglantziotis 1 · 0⤊ 0⤋
avg acc = (Vf-Vi)/t
avg acc = (10m/s-(-8m/s))/0.012 s
avg acc = 1500 m/s^2
2007-08-06 18:30:59 · answer #4 · answered by Stephen D 2 · 2⤊ 0⤋