An object moves across a straight line so that after t seconds, is distance from its starting point is: D(t)=t^3 - 12t^2 + 100t + 12 meters. Find the acceleration of the object after 3 seconds.
I believe acceleration is calculated by taking the second derivative.
D'(t) = 3t^2 - 24t + 100
D"(t) = 6t - 24
D"(3) = 18 - 24
D"(3) = -6
Hmmm...I have my doubts because it's a negative number.
You did everything correct, D"(t) is the acceleration. It happens to be decelerating at t=3 and at t=4 the acclereation turns positive.
Observe that the velocity is still positive at t=3.
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