Question

A pont P is moving along the x-axis. At any time t, the location of P on the x-axis is described by x=t^3-4t^2+3t. Determine the instantaneous acceleration at time t=5 of the point P.

Answer 1

x = t^3 - 4t^2 + 3t

we want to find the second derivative (first derivative of positive is velocity, second is acceleration)

thus

x' = 3t^2 - 4(2t^1) + 3(1)
= 3t^2 - 8t + 3

x'' = 3(2t^1) - 8(1) + 0
= 6t - 8

thus x'' = 6t - 8

so if we want the instantaneous acceleration at time t (in this case t = 5) just plug in 5 in the second derivative fn:

x'' = 6(5) - 8
= 22 m/s^2

Answer 2

can u help with this one? http://openstudy.com/users/geica#/updates/4f431b05e4b065f388dc8768