Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
Answer:
A works to find magnitude. The leading negative sign gives us the positive magnitude after the correct velocity is found without it.
Answer:
a = 0.01m/s²
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
a = (V_f-V_0)/t
a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))
a = 0.01m/s²
Answer:
<em>1. c. Same in both</em>
<em>2. a. Case 1</em>
<em></em>
Explanation:
1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.
2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.
