Course:MATH102/Question Challenge/2008 December Q5

Question

A particle’s motion is described by $y=t^{3}-6t^{2}+9t$ where y(t) is the displacement (in metres) t is time (in seconds) and $0\leq t\leq 4$ seconds.

(a) During this time interval, when is the particle furthest from its initial position ?

(b) During this time interval, what is the greatest speed of the particle?

(b) What is the total distance (including both forward and backward directions) that the particle has travelled during this time interval?

Hints

Hint
add your hints here

Solutions

Solution

(a) At time $t=0$, the particle is at the origin, and so $y(t)$ represents the position relative to the origin (and the displacement from the initial position). We are given that $y(t)=(t^{3}-6t^{2}+9t)$ To find the time when the particle is furthest away, we look for a maximum $y(t)$. First, we look for critical points. Setting the first derivative to zero, and solving for $t$: velocity $=y'(t)=3t^{2}-12t+9=0$ $0=t^{2}-4t+3=(t-1)(t-3)$ $t=1,3$ The critical points are at the times $t=1,3$ seconds. To find out what kind of critical point, we check the sign of the second derivative, at each of these points : acceleration =$y''(t)=6t-12$ $y''(1)=6(1)-12=-6<0$, so $t=1$ is a local maximum. $y''(3)=6(3)-12=6>0$, so $t=3$ is a local minimum. This shows that the particle is at maximum displacement from its initial position when $t=1$, and at minimum displacement from its initial position when $t=3$ Since the function is defined on a closed interval so $0\leq t\leq 4$, we need to check the endpoints. We calculate so $y(t)$ at the endpoints and at the local maximum, and find: $y(0)=0,\quad y(1)=4,\quad y(4)=4$ hence, the answer to (a) is $t=1,4$ (b) To find when the particle has maximum speed we will find the maximum of the velocity (first derivative) by setting its derivative (the second derivative, which is also the acceleration) equal to zero and solving for $t$: acceleration =$y''(t)=6t-12$ $0=6t-12$ $6t=12$ $t=2$ To check that this produces a maximum velocity and not a minimum velocity we will check the sign of the third derivative: $y'''(t)=6>0$ so $t=2$ produces a minimum, not maximum speed. We next check the endpoints: At $t=0$, velocity$=y'(0)=3(0)^{2}-12(0)+9=9$, At $t=4$, velocity$=y'(4)=3(4)^{2}-12(4)+9=9$. This shows that the particle has two maximum speeds at $t=0$ and $t=4$. (c) The direction changes at $t=1$ and at $t=3$. To find the total distance the particle has moved in both directions, we will find the difference in values from $t=0$ to the first critical point at $t=1$, and then the difference to the next critical point at $t=3$ and to the final endpoint at $t=4$. $y(0)=0$ $y(1)=(1)^{3}-6(1)^{2}+9(1)=4$ $y(3)=(3)^{3}-6(3)^{2}+9(3)=0$ $y(4)=(4)^{3}-6(4)^{2}+9(4)=4$ So the total distance traveled by the particle over the domain $0\leq t\leq 4$ is given by: $d=|y(1)-y(0)|+|y(3)-y(1)|+|y(4)-y(3)|$ $d=(4-0)+(4-0)+(4-0)$ $d=12$ Part a) The particle is furthest from its initial position when $t=1$ second. Part b) The particle has a maximum speed at two times within the domain of $0\leq t\leq 4$ at the endpoints $t=0$ and $t=4$ Part c) The total distance traveled by the particle both forward and backward is $12$ meters