Maximum Velocity

• Jan 16th 2011, 08:29 AM
zhangxupage2
Maximum Velocity
Hi everyone. I'm currently working on a velocity problem. Here's what I did:

A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t)=12t^2-36t+15. At t = 1, the particle is at the origin

Find the maximum velocity of the particle for 0≤t≤2.

V'=24t-36
24t-36=0
6(4t-6)=0
t=3/2

I drew a sign chart, and I found that 3/2 is actually the minimum. How can I get the maximum velocity?

Thanks a lot.
• Jan 16th 2011, 08:41 AM
Unknown008
You might want to make a sketch of that vt graph to help you.

$v(t)=12t^2-36t+15 = 3(4t^2 - 12t +5) = 3(2t - 1)(2t - 5)$

The x intercepts are at t = 0.5 and t = 2.5

Draw the graph and you'll see that the maximum will occur at t = 0.
• Jan 16th 2011, 06:46 PM
zhangxupage2
Thanks a lot.
• Jan 16th 2011, 07:54 PM
Soroban
Hello, zhangxupage2!

Quote:

$\text{A particle moves on the }x\text{-axis so that its velocity at any time }t \ge 0$
$\text{is given by: }\: v(t)\:=\:12t^2-36t+15$
$\text{At }t = 1\text{, the particle is at the origin.}$
$\text{Find the maximum velocity of the particle for }0 \le t \le 2.$

$v' \:=\:24t-36$
$24t-36\:=\:0 \quad\Rightarrow\quad t \,=\,\frac{3}{2}$

$\text{I drew a sign chart, and I found that }\frac{3}{2}\text{ is actually the minimum.}$ . Right!
$\text{How can I get the maximum velocity?}$

We know that the velocity function is a parabola.

And you found its vertex (minimum velocoty).

. . $f\left(\frac{3}{2}\right) \:=\:12(\frac{3}{2})^2 - 36(\frac{3}{2}) + 15 \:=\:-12$

So test the endpoints of the given interval.

. . $v(0) \:=\:12(0^2) - 26(0) + 15 \:=\: 15$

. . $v(2) \:=\:12(2^2) - 36(2) + 15 \:=\:-9$

Therefore . . .