31–32. Velocity functions A projectile is fired vertically upw...

Question

31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).

a. For the following functions s(t), find the instantaneous velocity function v(t). (Recall that the velocity function v is the derivative of the position function s.)

s(t)= −16t²+100t

Accepted Answer

Welcome back, everyone. In this problem, we want to find the instantaneous velocity function V of T if the rocket is launched vertically and its altitude in meters above the ground after T seconds is given by the function A of T equals 44 T2 + 60T. A says ZFT is -8 T + 60, B it's -40 + 60, C-8 T + 50, and D-4 T + 50. Now, if we are going to figure out the instantaneous velocity function VFT, then we have to ensure that we define the function AFT, which is our altitude function. And we already know it equals -402 plus 60T. Now recall that the instantaneous velocity function, OK, recall that V of T is the derivative of the altitude function. In other words, it's A of T. So if we differentiate our altitude function, then we should be able to find our instantaneous velocity function. So this means then that A of T is equal to the derivative with respect to T sorry V of T rather. My apologies. V of T is equal to the derivative with respect to T of our altitude function -402 + 60T. Now we can differentiate by using the power rule. The derivative of -40 squared is negative8T. We bring down the power 2 and we multiply it by -4 and then subtract 1 from the power. So that's how we get -8 T. and the derivative of 60T is 60. Therefore, V of T. Is equal to -8T plus 60. A is the correct answer. Thanks for watching everyone. I hope this video helped.

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