Consider a sports car which accelerates from 0 ft/sec to 88 ft/sec in 5 seconds (88 ft/sec = 60mph).The car's velocity is given in the table below. $\begin{array}{cccccc}
t & 0 & 1 & 2 & 3 & 4 &5\
V(t) & 0 & 30 & 52 & 68 & 80&88
\end{array}$ In which time interval is the average acceleration smallest?
A)Between t = 4 and t = 5.
B)Between t = 3 and t = 4.
C)Between t = 2 and t = 3.
D)Between t = 1 and t = 2.
E)Between t = 0 and t = 1.

Question

Quizplus · Accepted Answer

The average acceleration is calculated as the change in velocity divided by the change in time. Between t = 4 and t = 5, the change in velocity is $88 - 80 = 8$ ft/sec over 1 second, which is the smallest change in velocity per unit time compared to the other intervals provided.

Consider a Sports Car Which Accelerates from 0 Ft/sec to 88