- #1
vysero
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If a particle's position is given by x = 4-12t+3t^2 (where t is in seconds and x is in meters).
a) What is the velocity at t = 1s?
Ok, so I have an answer:
v = dx/dt = -12 + 6t
At t = 1, v = -12 + 6(1) = -6 m/s
but my problem is I want to see the steps of using the formula v = dx/dt in order to achieve -12 + 6t...
I am in phx with calc and calc is only a co-requisite for this class so I am taking it while I am taking physics but as you can see calc is a little behind. Where just now learning limits in calc and I was hoping someone could help me figure this out.
a) What is the velocity at t = 1s?
Ok, so I have an answer:
v = dx/dt = -12 + 6t
At t = 1, v = -12 + 6(1) = -6 m/s
but my problem is I want to see the steps of using the formula v = dx/dt in order to achieve -12 + 6t...
I am in phx with calc and calc is only a co-requisite for this class so I am taking it while I am taking physics but as you can see calc is a little behind. Where just now learning limits in calc and I was hoping someone could help me figure this out.