Question

A particle moves with velocity v(t) = t^2 -t measured in feet per second. How far did the particle travel
from t = 0 seconds to t = 3 seconds?

Answer 1

Is this a calc question?

Answer 2

yeah

Answer 3

integrate

Answer 4

i did integrate, but it's wrong

Answer 5

distance = velocity*time
so its the area under the curve v(t) from 0 to 3
\[\int\limits_{0}^{3}t^{2}-t = \frac{t^{3}}{3}-\frac{t^{2}}{2} from0->3\]

Answer 6

so is that 4.5. coz that's what i got eventually

Answer 7

yeah i get 4.5

Answer 8

ok, so then my teacher was wrong :) thx again dumbcow!

Answer 9

thx unklerhaukus and animalAin

Answer 10

Integrate from zero to three.\[\int\limits( t^2 -t)dt=t^3/3-t^2/2+C\]Evaluate at 3 and zero to get 9/2.

Answer 11

thx animalain

Answer 12

\[v(t)=\frac{\text{d} x}{\text {d}t} = t^2 -t\]
\[x=\int_0^3(t^2-t)\text{d}t \]\[={t^3\over 3} -{t^2\over2} |_0^3\]\[={3^3\over 3} -{3^2\over2} |_0^3\]\[=(9-{9\over2})-0\]\[={9\over2}\] so the particle has moved \[4.5 \text{ ft}\]

Answer 13

if you want the \(\LaTeX\) to look nice try

\[=\left.\frac{t^3}{3} -\frac{t^2}{2} \right|_{0}^{3}\]

=\left.\frac{t^3}{3} -\frac{t^2}{2} \right|_{0}^{3}