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Question 1168368: 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?
A. 5.5
B. 6.6
C. 7.7
D. 8.5
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 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?
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d(t) = 1.1t^2
d(2) = 4.4 feet
d(5) = 27.5 feet
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d(5) - d(2) = 23.1 feet
23.1/3 = 7.7 ft/sec
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