Tag Archives: acceleration vector

College Physics by Openstax Chapter 3 Problem 24

Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

Figure 3.61

Solution:

The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

Vectorx-componenty-component
A40\:\cos 60^{\circ} =20\:\text{km} 40\:\sin 60^{\circ} =34.6410\:\text{km}
B 30\:\cos 15^{\circ} =28.9778\:\text{km} 30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum 48.9778\: \text{km} 42.4056 \:\text{km}

The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. These are also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

\begin{align*}
\text{R} & = \sqrt{\left(48.9778\:\text{km}\right)^2+\left(42.4056\:\text{km}\right)^2} \\
\text{R} & = 64.8\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

The direction of the resultant is calculated as follows:

\begin{align*}
\theta & =\tan ^{-1}\left(\frac{42.4056}{48.9778}\right) \\
\theta & =40.9^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the pilot’s resultant displacement is about 64.8 km directed 40.9° North of East from the starting island.

Discussion:

If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up traveling southeast.

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College Physics by Openstax Chapter 3 Problem 23

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?

Solution:

Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

VectorX-ComponentY-Component
(1) -2.5\:\cos 45^{\circ} =-1.7678\:\text{km} +2.5\:\sin 45^{\circ} =+1.7678\:\text{km}
(2) +4.70\:\cos 60^{\circ} =+2.3500\:\text{km} -4.70\:\sin 60^{\circ} =-4.0703\:\text{km}
(3) -1.30\:\cos 25^{\circ} =-1.1782\:\text{km} -1.30\:\sin 25^{\circ} =-0.5494\:\text{km}
(4) +5.1000\:\text{km} 0
(5) +1.70\:\sin 5^{\circ} =+0.1482\:\text{km} +1.70\:\cos 5^{\circ} =+1.6935\:\text{km}
(6) -7.20\:\cos 55^{\circ} =-4.1298\:\text{km} -7.20\:\sin 55^{\circ} =-5.8979\:\text{km}
(7) +2.80\:\cos 10^{\circ} =+2.7575\:\text{km} +2.80\:\sin 10^{\circ} =+0.4862\:\text{km}
Sum 3.2799\:\text{km} -6.5701\:\text{km}

The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

\begin{align*}
\text{R} & =\sqrt{\left(3.2799\:\text{km}\right)^2+\left(-6.5701\:\text{km}\right)^2} \\
\text{R} & =7.34\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

The direction of the resultant is calculated as follows:

\begin{align*}
\theta & =\tan ^{-1}\left(\frac{6.5701}{3.2799}\right) \\
\theta & =63.47^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

Therefore, Gilligan is about 7.34 km directed 63.47° South of East from the starting island.

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College Physics by Openstax Chapter 2 Problem 36

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of  0.150 m/s2 as it goes through.  The station is 210 m long.

(a) How long is the nose of the train in the station?

(b) How fast is it going when the nose leaves the station?

(c) If the train is 130 m long, when does the end of the train leave the station?

(d) What is the velocity of the end of the train as it leaves?

Solution:

Part A

We are given the following: v_0=22.0\:\text{m/s}; a=-0.150\:\text{m/s}^2; and \Delta x=210\:\text{m}

We are required to solve for time, t. We are going to use the formula

\Delta x=v_0t+\frac{1}{2}at^2

Substituting the given values, we have

\begin{align*}
\Delta x & =v_0t+\frac{1}{2}at^2 \\
210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2
\end{align*}

If we simplify and rearrange the terms into a general quadratic equation, we have

0.075t^2-0.22t+210=0

Solve for t using the quadratic formula. We are given a=0.075;\:b=-22t;\:c=210.

\begin{align*}
t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\
t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\
\end{align*}

There are two values of t that can satisfy the quadratic equation.

t=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

Discard t=283.46 \ \text{s} as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve v_f in the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

So we have

\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\
v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

We are given the same values as in the previous parts, except that for the value of \Delta x since we should incorporate the length of the train. For the distance, \Delta x, we have

\begin{align*}
\Delta x & =210\:\text{m}+130\:\text{m} \\
\Delta x & = 340 \ \text{m}
\end{align*}

To solve for time t, we are going to use the formula

\Delta x=v_ot+\frac{1}{2}at^2

If we rearrange the formula into a general quadratic equation and solve for t using the quadratic formula, we come up with

t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\
t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\
t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

We have the same given values. We are going to solve for v_f in the equation

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\
v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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