An object starting from rest moves along a straight line such that, after t seconds, its distance from its starting point is S = 4t2 + 20 (t2 -1) meters. Find its acceleration in m/s2 after 4 seconds.
Added by Traci H.
Step 1
So, we have: v(t) = \frac{dS}{dt} = \frac{d}{dt}(4t^2 + 20(t^2 - 1)) Now, let's find the derivative: v(t) = 8t + 40t = 48t Show more…
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