The position function of an object is x(t) = -t^3 + 12t^2 - 27t+ 34

Indicate if the particle is speeding up or slowing down at t = 4

Question

The position function of an object is x(t) = -t^3 + 12t^2 - 27t+ 34

Indicate if the particle is speeding up or slowing down at t = 4

dumbcow · Answer

you need to find its acceleration
acceleration function is 2nd derivative of position function

anonymous · Answer

Yeah I know I'm supposed to compare the signs of the velocity and acceleration functions at that time

v(t) = -3t^2 + 24t - 27
v(4) = 21

a(t) = -6t + 24
a(4) = 0

uhhhh do 21 and 0 have the same sign? Am I doing something wron?

dumbcow · Answer

no you are correct
at t=4... the speed is 21 and acceleration is 0
this means its going a constant speed, its not speeding up or slowing down

dumbcow · Answer

a(t) >0  --> speeding up

a(t) < 0  --> slowing down

anonymous · Answer

Oh okay that makes sense...thanks so much! :)

dumbcow · Answer

yw