<PRE><B>Question 1168368</B>
A car leaving a stop sign accelerates constantly from a speed of 0 feet per
second to reach a speed of 44 feet per second. The distance of the car from the stop sign,d, in feet, at time t, in seconds, can be found using the equation d=1.1t^2 .
What is the average speed of the car, in feet per second, between
t = 2 and t = 5?
--------------
d(t) = 1.1t^2
d(2) = 4.4 feet
d(5) = 27.5 feet
----
d(5) - d(2) = 23.1 feet
23.1/3 = 7.7 ft/sec</PRE>