A Speed and Position question?????

Question

The position function of a particle is given by r(t) = ( 10t^2, 8t, 10t^2- 160t). When is the speed minimum? I am getting t = 2.

Answer 1

since u have the position. u can take the derivative of the fuctions to get speed. n for minimum value, just input the lowest time value in the derived functions.

be sure to know if the question is asking for the final magnitude or scalar form.

Answer 2

r(t) = ( 10t^2, 8t, 10t^2- 160t)

v(t)=r'(t)=(20t,8,20t-160)

|v(t)|=√[(20t)^2+8^2+(20t-160)^2]

|v(t)|=√(800t^2-6400t+25664)

Max or min speed is at |v(t)|'=0

|v(t)|'=(1600t-6400)/[2√(800t^2-6400t+... ==>

1600t-6400=0

t=4 ==> min|v(t)|=|v(4)|

It's obviously a min speed, because x-  and z-components are proportional to t^2,

while y-comp is proportional to t.