Question 6

A particle starting at the origin, moves along the $x$-axis with velocity $v$ meters per second at time $0 \le t \le 5$ seconds given by the formula:

$$v = t - \frac{8}{t + 2}.$$

• Plot the velocity as a function of time and determine the intervals during which the particle is moving forward and the intervals during which the particle is moving backwards.
  
  When $t = 0$ we have $v = 0 - \frac{8}{2} = - 4$, so initially the particle is going backwards.
  When $t = 5$, we have $v = 5 - \frac{8}{7} = \frac{27}{7}$, so at the end the particel is going forwards.
  The function $v$ is continuous on $[0, 5]$, so by the intermediate value theorem, $v$ must be zero somewhere in the interval $(0, 5)$.
  Putting $v = 0$, we get:
  $$0 = t - \frac{8}{t + 2},$$
  $$0 = t(t + 2) - 8,$$
  $$0 = t^2 + 2t - 8 = (t + 4)(t - 2),$$
  $$t = 2, \hspace{3pt} \textrm{or} \hspace{3pt} t = - 4.$$
  So in the given interval, $v$ vanishes only when $t = 2$.
  Then for $t < 2$, the particle is moving backwards and for $t > 2$, the particle is moving forwards.
  
  Plotting $v$ against $t$, we see that it is almost a straight line with $v$ increasing with time, slightly curving downwards:
  • We have the slope as $${\frac{dv}{dt} = 1 + 8(t + 2)^{-2}}$$ which is positive and a little more than $1$ in the given interval.
  • Also we have $${\frac{d^2 v}{dt^2} = - 16(t + 2)^{-3}}$$, which is negative in $[0, 5]$, so the curve is concave down in the given interval.
• Find the displacement of the particle from its initial position after five seconds.
  If the position of the particle is $x$ at time $t$, we have: $${\frac{dx}{dt} = v}$$, so, by FTC, we have the displacement in meters of the particle from its initial position as:
  $$x(5) - x(0) = \int_0^5 \frac{dx}{dt} dt = \int_0^5 v dt$$
  $$= \int_0^5 (t - \frac{8}{t + 2} ) dt$$
  $$= [\frac{t^2}{2} - 8 \ln(t + 2)]_0^5$$
  $$= \frac{25}{2} - 8\ln(7) - ( 0 - 8\ln(2))$$
  $$= \frac{25}{2} - 8\ln(\frac{7}{2}) = 2.47789625.$$

• Find the total distance traveled by the particle in the first five seconds.
  If $s$ is the distance travelled to time $t$, we have $${ \frac{ds}{dt} = \vert v\vert}$$, so by FTC, we have that the total distance travelled during the interval $[0, 5]$ as:
  $$\int_{t =0}^{t = 5} ds = \int_0^5 \frac{ds}{dt} dt$$
  $$= \int_0^5 \vert v\vert dt = \int_0^2 (- v) dt + \int_2^5 v dt$$
  $$= \int_2^5 (t - \frac{8}{t + 2} ) dt - \int_2^5 (t - \frac{8}{t + 2} ) dt$$
  $$= [\frac{t^2}{2} - 8 \ln(t + 2)]_2^5 - [\frac{t^2}{2} - 8 \ln(t + 2)]_0^2$$
  $$= \frac{25}{2} - 8\ln(7) - ( \frac{4}{2} - 8\ln(4)) - ( ( \frac{4}{2} - 8\ln(4)) - (0 - 8\ln(2)))$$
  $$= \frac{25}{2} - 4 - 8\ln(7) + 16\ln(4) - 8 \ln(2)$$
  $$= \frac{17}{2} + 24\ln(2) - 8\ln(7) = 9.568251141.$$