Motion in a Straight Line (College Board AP® Calculus BC): Revision Note

A particle is moving along the $x$-axis. At time $t$ seconds, the velocity of the particle is given by $v\left(t\right)={\left(t^2-5\right)}^3$ inches per second, and the acceleration of the particle is given by $a\left(t\right)=6t{\left(t^2-5\right)}^2$ inches per second per second.

Is the speed of the particle increasing, decreasing, or neither at time $t=2$ seconds? Give a reason for your answer.

Answer:

Find the values of the velocity and acceleration by substituting $t=2$ into the expressions

$v\left(2\right)={\left(2^2-5\right)}^3=-1$ inches per second

$a\left(2\right)=6\left(2\right){\left(2^2-5\right)}^2=12$ inches per second per second

Consider the signs

$$\begin{gathered} v\left(2\right)<0 \\ a\left(2\right)>0 \end{gathered}$$

The speed is decreasing at $t=2$ seconds because the signs are different for the velocity and acceleration

A particle moves along the $x$-axis. The velocity of the particle at time $t$ is given by a differentiable function $v$, where $v\left(t\right)$ is measured in meters per second and $t$ is measured in seconds. Selected values of $v\left(t\right)$ are given in the table above.

(a) Justify why there must be at least two times, for $0<t<8$, when the particle is at rest.

(b) Approximate $a\left(5\right)$, the acceleration of the particle at time $t=5$ seconds. Show the computations that lead to your answer, and indicate units of measure.

Answer:

(a)

The particle is at rest when $v\left(t\right)=0$

You need to use the intermediate value theorem

First, state why the function is continuous

$v\left(t\right)$ is differentiable ⇒ $v\left(t\right)$ is continuous

Then, find two pairs of values of $t$ for which the values of $v\left(t\right)$ have different signs

• Note there are many possible pairs

$v\left(2\right)=2>0$ and $v\left(4\right)=-3<0$

$v\left(6\right)=-1<0$ and $v\left(8\right)=4>0$

Therefore, by the intermediate value theorem, there is at least one value $t_1$ for $2<t_1<4$ and at least one value $t_2$ for $6<t<8$ such that $v\left(t_1\right)=v\left(t_2\right)=0$

Therefore, there are at least two times when the particle is at rest

(b)

Acceleration is the rate of change of the velocity

$a\left(5\right)=v\text{'}\left(5\right)$

Find the average rate of change using the two closest values to $t=5$

$\begin{array}{rcl}v\text{'}\left(5\right)& \approx & \frac{v\left(6\right)-v\left(4\right)}{6-4}\\ & =& \frac{-1-\left(-3\right)}{2}\\ & =& 1\end{array}$

Include the units

The acceleration of the particle at time $t=5$ seconds is approximately 1 meter per second per second

The displacement from the origin of a particle, $P$, as it travels along the $x$-axis is given by $s_P=3\sin \left(\frac{1}{2}t\right)$.

The displacement from the origin of a second particle, $Q$, as it travels along the $x$-axis is given by $s_Q=\frac{1}{8}{t}^3-\frac{7}{4}{t}^2+5t$.

$s_P$ and $s_Q$ are measured in meters and $t$ is measured in seconds for $0\le t\le 10$.

(a) Determine which particle is furthest from the origin at $t=5$.

(b) At $t=5$, determine if the particles are moving closer together or further apart. Explain your reasoning in your working.

(c) At $t=5$, determine which particle has the greatest magnitude of acceleration.

Answer:

(a)

Find the displacement of each particle at $t=5$

$s_P=3\sin \left(\frac{1}{2}·5\right)=1.795416...$ meters

$s_Q=\frac{1}{8}{\left(5\right)}^3-\frac{7}{4}{\left(5\right)}^2+5\left(5\right)=-3.125$ meters

$P$ is 1.795... meters away from the origin in the positive direction, while $Q$ is 3.125 meters away from the origin in the negative direction

At $t=5$ particle $Q$ is the furthest from the origin.

(b)

This question is about the direction of motion, so we need to find the velocity

Find expressions for the velocities by differentiating the expressions for the displacements

$v_P=\frac{d{s}_P}{dt}=\frac{3}{2}\cos \left(\frac{1}{2}t\right)$

$v_Q=\frac{d{s}_Q}{dt}=\frac{3}{8}{t}^2-\frac{7}{2}t+5$

Find the velocities when $t=5$

$v_P=\frac{3}{2}\cos \left(\frac{1}{2}·5\right)=-1.201715...$ meters per second

$v_Q=\frac{3}{8}{\left(5\right)}^2-\frac{7}{2}\left(5\right)+5=-3.125$ meters per second

Consider the positions and velocities of the two particles

$P$ is on the positive side (right) of the $x$ axis (1.795... meters), moving with a negative velocity, so back towards the origin (moving to the left)

$Q$ is on the negative side (left) of the $x$ axis (-3.125... meters), moving with a negative velocity, so further away from the origin (moving to the left)

$$\begin{gathered} \stackrel{3.125}{\leftarrow }\stackrel{1.2017}{\leftarrow } \\ \overline{)Q}\overline{)P} \end{gathered}$$

So both particles are moving to the left (negative $x$ direction) but $Q$ is moving faster, and its position is further to the left, so $Q$ is "escaping" from $P$

Therefore at $t=5$, the particles are moving further apart

(c)

Find expressions for the accelerations by differentiating the expressions for the velocities

$a_P=\frac{d{v}_P}{dt}=-\frac{3}{4}\sin \left(\frac{1}{2}t\right)$

$a_Q=\frac{d{v}_Q}{dt}=\frac{3}{4}t-\frac{7}{2}$

Find the acceleration of each when $t=5$

$a_P=-\frac{3}{4}\sin \left(\frac{1}{2}·5\right)=-0.448854...$ meters per second squared

$a_Q=\frac{3}{4}\left(5\right)-\frac{7}{2}=0.25$ meters per second squared

Whilst the acceleration of $Q$ is the largest (as it is positive whereas $P$'s is negative), it is $P$ whose acceleration has the greatest absolute value (magnitude)

At $t=5$, particle $P$ has the greatest magnitude of acceleration.