Question

Find the average velocity for the first three seconds

Answer 1

Another one I'm stuck on...
\[x(t) = 3t - e (12 - 4t)\]
The 12-4t should all be an exponent on the e

Answer 2

So... I know that x'(t) = 3 - e^(12-4t)(-4)
But... what exactly do they want me to do?

Answer 3

This is easier than you're making it.

Average velocity is displacement over time.

Displacement is change in position.

Find the position at t=3 (ie find x(3) ) find the position at t=0 (ie x(0) )

Subtract them, divide by time of 3 seconds.

\[\large \frac{ x(3) - x(0) }{ 3 }\]

Answer 4

xD wow. I took x'(3) + x'(0) divided by 3
I'm getting
\[\frac{ 8 + e(12) }{ 3 }\]
with 12 being the exponent..

Answer 5

What you did, is change in velocity divided by time. That's average acceleration.

Answer 6

Oh. :P
Is 8 + e^12 over 3 right?

Answer 7

\[\large x(t) = 3t - e ^{12 - 4t}\]
\[\Large \frac{ 3(3) - e ^{12 - 4(3)} - (3(0) - e ^{12 - 4(0)}) }{ 3 } = \frac{ 8 + e ^{12 } }{ 3 } \]

Answer 8

Thanks :)