Question

Let s=30/(t^2+12) be the position function of a particle moving along a coordinate line, where s is in feet and t is in seconds.

(a) Find the maximum speed of the particle for t>=0 . If appropriate, leave your answer in radical form.

Speed (ft/sec): ?

(b) Find the direction of the particle when it has its maximum speed.

Answer 1

If t is >= 0, plug in 0 in the equaion and you'll get :

s(0) = 30/12 ft/sec

I guess ^_^

Answer 2

I don't think that is the correct, answer. Give me a few minutes, I will do this problem.

Answer 3

Ok, to obtain the maximum speed you would have to take the second derivative of s(t)
So, s''(t) = 0 will give you the maximum speed,
s'(t)=0 will give you where the speed is zero not maximum. (this is why the above answer is incorrect).

s''(t) = 0 solving for t will be +2 and -2
choosing t>=0 then t = +2
therefore plug it back into the velocity equation s'(t) which is going to be at time t=2
s'(2)= -15/32 Feet/Second, that would be the velocity. speed would just be the absolute value, so speed at time t=2 is 15/32

b) the direction will be in the negative direction since the velocity is negative, and the initially we assume the object to be at coordinate (0,5/2) given from s(0)

Answer 4

for your reference,
s'[t]=-((60 t)/(12 + t^2)^2)
s''[t]=[(240 t^2)/(12 + t^2)^3] - [ 60/(12 + t^2)^2]