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sonic pwr
Registered User Posts: **769** Member

a particle moves on the x-axis in such a way that its position at time t is given by x=(2t-1)((t-1)^2)

i know that speed is absolute value of velocity and when velocity is 0 t= 2/3 and 1

so during the interval (2/3 , 1) when is the speed at a maximum??

i know that speed is absolute value of velocity and when velocity is 0 t= 2/3 and 1

so during the interval (2/3 , 1) when is the speed at a maximum??

Post edited by sonic pwr on

## Replies to: how to find maximum speed???

183Junior Memberv(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

769Member157Junior Memberabs(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.

769Member157Junior MemberSo, you see, with the first notation you actually have extremas at the endpoints. Hence why you needed to check if speed was greatest there.

183Junior Member157Junior MemberBut c'mon, who is never been guilty of not reading a question completely? =P

769Member157Junior MemberSo you move to the left at 2/3<t<1 or (2/3,1).