Question

HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.

Answer 1

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when $t=0$ and $t=3$, and because the velocity function is continuous, you need only check the sign of $v(t)$ for values on the intervals (0, 3) and (3, 6).

We have, for instance $v(1)\approx-0.91$ and $v(4)\approx0.91>0$, which means the particle is moving the positive direction for $3$, or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

$\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt$

which follows from the definition of absolute value. In particular, if $x$ is negative, then $|x|=-x$.

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so $a(t)$ is the derivative of $v(t)$:

$a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)$

Compute the acceleration at $t=4$ seconds:

$a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}$

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)