Question

The velocity given by v=2t^2-12t+8. find acceleration of particle after 4 sec, and find the minimum velocity?

Answer 1

Have you considered dv/dt = a(t), the acceleration?

Answer 2

Find the derivative of the velocity function, then evaluate it at 4.

Answer 3

I founded but the question ask about minimum Velocity? how to find it?

Answer 4

how did u get it

Answer 5

you plug in zero in which equation???
and how to find max velocity?

Answer 6

I don't think so

Answer 7

the function of the velocity is v = 2t^2 -12t+8 which is an upward parabola. Therefore, the minimum velocity is at its vertex, which is t = -b /2a = -(-12)/2*2 = 3.

Answer 8

ok, this is t, but then plug in 3 which equation??

Answer 9

yes.

Answer 10

plug back to the original one

Answer 11

v = 2 t^2..........

Answer 12

will get -10,, is that right!

Answer 13

how about the max velocity!

Answer 14

I don't know how to get max if I don't have interval of time.

Answer 15

what I understood, I should plug in zero in velocity equation!

Answer 16

to get max velocity!!

Answer 17

https://www.desmos.com/calculator this is your graph, show me how to get the max?

Answer 18

I don't know how to graph?

Answer 19

just do as I did, go to that site, type your function from the left hand side box and it automatically draw it for you. Of course, It is just for confirm our calculation.

Answer 20

There is no Maximum Velocity. The velocity function is unbounded. Please stop looking for the maximum velocity. You should know this from the nature of the parabola. That's why you studied Analytic Geometry prior to studying the calculus.

Please listen to Loser66. Without a specific and finite interval of time, there is no maximum velocity.

Answer 21

If you see the velocity function, it is the equation of a parabola opening up.