well you have the equation x=(2t-1)((t-1)^2)... so take the derivative of that to find eq for speed... first d second plus second d first...
v(t) = (2t - 1)(2t - 2) + ((t - 1)^2)(2)
v(t) = (4t^2 - 6t + 2) + (2t^2 - 4t +2)
v(t) = 6t^2 -10t +4
so then to find where speed is greatest, find the critical numbers (where deriv is 0 or undef). and i know i'm working with the velocity eq... so we'll count min's as well as max's to find "max speed"
v'(t) = 12t - 10
0 = 12t - 10
t = 5/6
so plug in 5/6 to velocity
v(5/6) = 6*(5/6)^2 - 10(5/6) - 4
..........= 25/6 - 50/6 + 24/6
..........= -1/6
take absolute value, and the max speed is 1/6 units/sec
i guess you didn't need to know what the max speed was, but i did it anyway. but the t value for the max is 5/6.
Is that an open or closed interval? If it's a close interval you must check the endpoints for speed.
abs(v(2/3)) = abs(v(1)) < abs(v(5/6))
K, answer for max speed still occurs at time = 5/6
The particle moves to the left when velocity is negative, so the entire interval, yes.
At t = 2/3 and t = 1, since velocity is zero, you are in the process of changing directions (you are not moving to the left).
So you move to the left at 2/3<t<1 or (2/3,1).
Replies to: how to find maximum speed???
v(t) = (2t - 1)(2t - 2) + ((t - 1)^2)(2)
v(t) = (4t^2 - 6t + 2) + (2t^2 - 4t +2)
v(t) = 6t^2 -10t +4
so then to find where speed is greatest, find the critical numbers (where deriv is 0 or undef). and i know i'm working with the velocity eq... so we'll count min's as well as max's to find "max speed"
v'(t) = 12t - 10
0 = 12t - 10
t = 5/6
so plug in 5/6 to velocity
v(5/6) = 6*(5/6)^2 - 10(5/6) - 4
..........= 25/6 - 50/6 + 24/6
..........= -1/6
take absolute value, and the max speed is 1/6 units/sec
i guess you didn't need to know what the max speed was, but i did it anyway. but the t value for the max is 5/6.
-will
abs(v(2/3)) = abs(v(1)) < abs(v(5/6))
K, answer for max speed still occurs at time = 5/6
The particle moves to the left when velocity is negative, so the entire interval, yes.
So, you see, with the first notation you actually have extremas at the endpoints. Hence why you needed to check if speed was greatest there.
But c'mon, who is never been guilty of not reading a question completely? =P
So you move to the left at 2/3<t<1 or (2/3,1).