document.write( "Question 1168368: A car leaving a stop sign accelerates constantly from a speed of 0 feet per
\n" ); document.write( "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 .
\n" ); document.write( "What is the average speed of the car, in feet per second, between
\n" ); document.write( "t = 2 and t = 5?
\n" ); document.write( "A. 5.5
\n" ); document.write( "B. 6.6
\n" ); document.write( "C. 7.7
\n" ); document.write( "D. 8.5
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #792952 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
A car leaving a stop sign accelerates constantly from a speed of 0 feet per
\n" ); document.write( "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 .
\n" ); document.write( "What is the average speed of the car, in feet per second, between
\n" ); document.write( "t = 2 and t = 5?
\n" ); document.write( "--------------
\n" ); document.write( "d(t) = 1.1t^2
\n" ); document.write( "d(2) = 4.4 feet
\n" ); document.write( "d(5) = 27.5 feet
\n" ); document.write( "----
\n" ); document.write( "d(5) - d(2) = 23.1 feet
\n" ); document.write( "23.1/3 = 7.7 ft/sec
\n" ); document.write( "
\n" );