(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).)

6 0

3 years ago

How can I solve this? 10+x = 5(1/5x + 2)

pshichka [43]

You use PEMDAS. Start by distributing the terms in the parentheses. The isolate the variables. To do that you subtract x from one side of the equation to the other side. You always do the other operation. If it is adding you subtract, if your multiplying your dividing and the other way around. Then you subtract 10 from each side. This equation has many solution because all the terms cancel out.

5 0

3 years ago