You might want to make a sketch of that vt graph to help you.
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.
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.