Question

vx = (5.9 + 1.1t + 7.1t^2) where t is in seconds. Find the average acceleration (in m/s^2) in the time interval t = 5.4 s to t = 6.7 s. 

          vx = (5.9 + 1.1t + 7.1t^2) where t is in seconds. Find the average acceleration (in m/s^2) in the time interval t = 5.4 s to t = 6.7 s.

Added by Alicia A.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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vx = (5.9 + 1.1t + 7.1t^2) where t is in seconds. Find the average acceleration (in m/s^2) in the time interval t = 5.4 s to t = 6.7 s.
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Transcript

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00:01 Without being given x of t or v of t or a of t, i can't quite answer this question, but i can tell you a basic formula.
00:11 So i'm going to start with v of t with average acceleration, at least specific to this problem.
00:19 So they gave you velocity.
00:22 You could find the average acceleration on the interval by doing v of six minus v of one over six minus one.
00:31 And if you wanted to, you could actually simplify, but i can't do anything else until i know what v of six or v of one is.
00:39 But i do know that six minus one is five.
00:42 Now a little subtle difference is what if they gave you acceleration and they asked for average acceleration? you can still do this by doing one over your upper bound minus lower bound six minus one, and then the integral from one to six of a of t dt...
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