just solve for t when v=0
v = 5.41 - 5.4t
looks like t=1
It does not stop decelerating. There is a constant acceleration of -5.4m/s^2
The car will move in the +x direction for 1 second, then stop and reverse direction. No conditions were given on what would make it then stop, but it would be back to its original position at time t=2.
A car is moving in the +x direction with the following position equation of motion: x = 5.41(m/s)t - 2.7(m/s^2)t^2.
(a) What is the initial velocity of the car?
5.41 m/s
(b) What is the acceleration of the car?
-5.4 m/s^2
(c) At what time does the car come to rest?
__________ s
(d) At what time does the car stop "decelerating?"
__________ s
NEED HELP WITH (C) AND (D)
1 answer
1 answer