Question

The velocity of a particle moving along the x-axis is: v(t)= -(t-1)(^2) + 2... for t between 0 and 6. if at t=0 the particle is at the origin, find the location of the particle at t=3

Answer 1

\[\textbf v(t) = \frac{\mathrm{d}\textbf x(t)}{\mathrm{d}t}\]
\[\int_{0}^{t≤6s} \textbf v(t) \mathrm{d}t=\int_{x_0=0}^{x}\mathrm{d}x\]
\[\begin{array}{l} \textbf x(3) & =\int_{0}^{3} \big(-(t-1)^2+2 \big) \ \mathrm{d}t
\\ & = \int_{0}^{3}(1+2t-t^2) \mathrm{d}t
\end{array}\]
Then integrate


Remember for the sake of completeness that the units on velocity are m/s, and the units on position are meters, so technically the units on each term of the integration expression is


\[\int_{0}^{t}(1m/s)+(2m/s^2)t-(1m/s^3)t^2) \mathrm{d}t\]
This doesn't change the numeric answer, but is good to think about.

Answer 2

Make sense? ^_^

Answer 3

yes thank you!

Answer 4

^_^