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<0` and
`v(4)\approx0.91>0`, which means the particle is moving the positive direction for
`3<t<6`, 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)=-\frac{\pi^2}9\cos\left(\frac{\pi t}3\right)`

Compute the acceleration at
`t=4` seconds:

`a(t)=(\pi^2)/(18)(\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).)