Question

A particle moves according to a law of motion \(s = f\left( t \right)\), \(t \ge 0\), where \(t\) is measured in seconds and \(s\) in feet.

(a) Find the velocity at time \(t\).

(b) What is the velocity after 1 second?

(c) When is the particle at rest?

(d) When is the particle moving in the positive direction?

(e) Find the total distance traveled during the first 6 seconds.

(f) Draw a diagram like Figure 2 to illustrate the motion of the particle.

(g) Find the acceleration at time t and after 1 second.

(h) Graph the position, velocity, and acceleration functions for \(0 \le t \le 6\).

(i) When is the particle speeding up? When is it slowing down?

4. \(f\left( t \right) = {t^2}{e^{ - t}}\)

Short answer

1. The velocity\(v\left( t \right)\)is \(v\left( t \right) = t{e^{ - t}}\left( { - t + 2} \right)\).
2. The velocity after 1 sec is \(1/e\,\,{\rm{ft/s}}\).
3. The particle is at rest at 0 s and 2 s.
4. The particle is moving in a positive direction in the intervals \(0 < t < 2\).
5. The total distance covered in 6 sec is \(0.99\)ft approximately.
6. The diagram is

1. The acceleration \(a\left( t \right)\) is \(a\left( t \right) = {e^{ - t}}\left( {{t^2} - 4t + 2} \right)\) and at 1 sec is \( - 1/e\,{\rm{ft/}}{{\rm{s}}^2}\)
2. The graph is

1. The particle is speeding up in the intervals \(2 < t < 3.4\), \(0 < t < 0.6\)and slowing down in the intervals \(0.6 < t < 2\),\(t > 3.4\).

Step-by-step solution

Find the velocity \(v\left( t \right)\)

The velocity of the motion function is obtained as \(v\left( t \right) = f'\left( t \right)\). Thus, the equation of velocity, obtained by differentiation, is given by

\(\begin{aligned}v\left( t \right) & = {t^2}\left( { - {e^{ - t}}} \right) + {e^{ - t}}\left( {2t} \right)\\ & = t{e^{ - t}}\left( { - t + 2} \right)\end{aligned}\)

Plug\(t = 1\)into \(v\left( t \right) = t{e^{ - t}}\left( { - t + 2} \right)\), to find the velocity after 1 second, as shown below:

\(\begin{aligned}v\left( 1 \right) & = \left( 1 \right){e^{ - 1}}\left( { - 1 + 2} \right)\\ & = \frac{1}{e}\,{\rm{ft/s}}\end{aligned}\)

The particle is at rest when velocity is 0. Plug\(v\left( t \right) = 0\) into \(v\left( t \right) = t{e^{ - t}}\left( { - t + 2} \right)\), and simplify to find the time \(t\) when the particle is at rest, as shown below:

\(\begin{aligned}t{e^{ - t}}\left( { - t + 2} \right) & = 0\\t & = 0,2\end{aligned}\)

The particle is moving in the positive direction if \(v\left( t \right) > 0\). Solving \(v\left( t \right) > 0\) to find time at which velocity is positive, as follows:

\(\begin{aligned}t{e^{ - t}}\left( { - t + 2} \right) > 0\\t\left( { - t + 2} \right) > 0\\0 < t < 2\end{aligned}\)

Estimate the distance

As the particle velocity changes sign at\(t = 2\)in the interval\(\left( {0,6} \right)\).So, the distance will also be calculated in the intervals\(\left( {0,2} \right)\)and\(\left( {2,6} \right)\)only, as shown below:

\(\begin{aligned}\left| {f\left( 2 \right) - f\left( 0 \right)} \right| & = \left| {4{e^{ - 2}} - 0} \right|\\ & = 4{e^{ - 2}}\end{aligned}\)

\(\begin{aligned}\left| {f\left( 6 \right) - f\left( 2 \right)} \right| & = \left| {36{e^{ - 6}} - 4{e^{ - 2}}} \right|\\ & = 4{e^{ - 2}} - 36{e^{ - 6}}\end{aligned}\)

Thus, the total distance traveled in 6 seconds is \(4{e^{ - 2}} + 4{e^{ - 2}} - 36{e^{ - 6}}\), that is, 0.99 ft. approximately.

Draw the graph showing motion of a particle

Based on information gathered, the motion of the particle is graphed as follows:

Find the acceleration \(a\left( t \right)\)

The acceleration of the motion function is obtained as \(a\left( t \right) = v'\left( t \right)\). Thus, the equation of acceleration, obtained by differentiation, is given by :

\(\begin{aligned}a\left( t \right) & = \left( {2t - {t^2}} \right)\left( { - {e^{ - t}}} \right) + {e^{ - t}}\left( {2 - 2t} \right)\\ & = {e^{ - t}}\left( { - \left( {2t - {t^2}} \right) + \left( {2 - 2t} \right)} \right)\\ & = {e^{ - t}}\left( {{t^2} - 4t + 2} \right)\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\).

Plug\(t = 1\)into \(a\left( t \right) = {e^{ - t}}\left( {{t^2} - 4t + 2} \right)\), to find the acceleration after 1 second, as shown below:

\(\begin{aligned}a\left( 1 \right) & = - {e^{ - 1}}\left( {1 - 4 + 2} \right)\\ & = - 1/e\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\)

Draw the graph of velocity, position, and acceleration from 0 to 6 sec

Using the equation of the position\(f\left( t \right)\), velocity \(v\left( t \right)\) and acceleration \(a\left( t \right)\), draw these equations within \(0 \le t \le 6\)sec.

Draw the graph of the functions\(f\left( t \right) = {t^2}{e^{ - t}}\),\(v\left( t \right) = t{e^{ - t}}\left( { - t + 2} \right)\)and\(a\left( t \right) = {e^{ - t}}\left( {{t^2} - 4t + 2} \right)\)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\({X^2}{e^{ - X}}\)in the\({Y_1}\)tab,\(X{e^{ - x}}\left( { - X + 2} \right)\)in the\({Y_2}\)tab, and\({e^{ - X}}\left( {{X^2} - 4X + 2} \right)\)in the\({Y_3}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the functions\(f\left( t \right) = {t^2}{e^{ - t}}\),\(v\left( t \right) = t{e^{ - t}}\left( { - t + 2} \right)\) and \(a\left( t \right) = {e^{ - t}}\left( {{t^2} - 4t + 2} \right)\) is shown below:

Comment on the speed of the particle

Set \(a\left( t \right) = 0\), and find time \(t\) at which acceleration changes sign as follows:

\(\begin{aligned}&{t^2} - 4t + 2 = 0\\t & = \frac{{4 \pm \sqrt 8 }}{2}\\ & = 2 \pm \sqrt 2 \\ & \approx 0.6,\,3.4\end{aligned}\)

The particle is speeding up when velocity and acceleration have the same sign. For the duration \(2 < t < 3.4\), velocity and acceleration are both negative. For the duration \(0 < t < 0.6\) velocity and acceleration are both positive.

The particle is slowing when velocity and acceleration have the opposite sign. For the duration \(0.6 < t < 2\),\(t > 3.4\), velocity and acceleration have opposite signs.