a driver traveling 20m/s at an acceleration of 4/3 m/s squared begins to brake with a deceleration of magnitude 3 m/s squared determine the time it takes the car to stop
+33661 You can use the constant acceleration formula: v = u + at
v is final velocity (0 m/s)
u is initial velocity (20 m/s)
a is acceleration (-4/3 m/s2: the minus sign shows it is decelerating)
t is the time (in seconds)
So:
0 = 20 - (4/3)*t
Rearrange:
(4/3)t = 20
t = 20*3/4
t = 15 seconds
+33661 You can use the constant acceleration formula: v = u + at
v is final velocity (0 m/s)
u is initial velocity (20 m/s)
a is acceleration (-4/3 m/s2: the minus sign shows it is decelerating)
t is the time (in seconds)
So:
0 = 20 - (4/3)*t
Rearrange:
(4/3)t = 20
t = 20*3/4
t = 15 seconds
