The position of an object as a function of time is given by x(t)=at^3-b(t)^2+ct-d, where a=3.6 m/s^3, b=5.0m/s^2, c=6.0m/s and d=7.0m
a) find the instantaneous acceleration at t=2.4s
b) find the average acceleration over the first 2.4 second

Question

The position of an object as a function of time is given by x(t)=at^3-b(t)^2+ct-d, where a=3.6 m/s^3, b=5.0m/s^2, c=6.0m/s and d=7.0m 
a) find the instantaneous acceleration at t=2.4s 
b) find the average acceleration over the first 2.4 second

anonymous · Answer

for a) i took the second derivative and plugged in the values and got 41m/s/s..is that correct

anonymous · Answer

Hopefully you know acceleration is the second derivative of position with respect to time.

anonymous · Answer

yeh

anonymous · Answer

but then i got stuck on the second part

anonymous · Answer

i know its its change in velocity/time too

anonymous · Answer

( V(2.4) -V(0) )/(2.4)

anonymous · Answer

but how do i do that

anonymous · Answer

differentiate once and plug '2.4' and '0' into V(t)

anonymous · Answer

find the difference between those two results, divide by 2.4.

anonymous · Answer

oh ok got it. thank u!

anonymous · Answer

sure!