Answer:
Find the time it took for the car to stop at 11.0m/s
V = deltax/t
t = 41.14/11.0 = 3.74s
Now find at what rate it was decelerating, so find the acceleration during that interval of time.
vf = vi + at
-11.0m/s = a3.74s
a = -2.94m/s^2
The acceleration is negative because is pulling the car towards its opposite direction to make it stop.
Now find how much time it would take for the car to stop at 28.0m/s but with the same acceleration, the car is the same so its acceleration to stop the car will remain the same.
vf = vi + at
0 = 28.0 - 2.94t
t = 9.52
Once the time is obtained, you can find the final position, xf, by plugging the time acceleration and velocity values.
xf = 0 + (28m/s)(9.52s) + 1/2(-2.94)(9.52s)^2
xf = 266.6m - 133.23m = 133m
Answer:
Explanation:
This problem can be solved with the conservation of the momentum.
If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.
Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.
If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.
HOPE THIS HELPS!!
I don't really know but I don't think you can... I can't explain it either. Sorry im no help