change in velocity/ change in time = (17.9-29.6) / 4.08
= -2.67 m/s^2
d = Vi t - (1/2) 2.67 * t^2
= 29.6(4.08) - 1.33 (4.08^2)
= 121 - 22.1
= 98.9 m
1) A car moving with an initial velocity of 29.6 m/s decreases to a velocity of 17.9 m/s over a time
interval of 4.08 seconds.
(a.) What is the average acceleration of the car during this time?
(b.) Assuming the acceleration is constant, how far did the car travel during this time?
1 answer
1 answer