Chapter 3
Physics for Scientists and Engineers with Modern Physics · 110 exercises
Problem 1
(I) A car is driven \(225 \mathrm{~km}\) west and then \(78 \mathrm{~km}\) southwest \(\left(45^{\circ}\right)\). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.
5 step solution
Problem 1
(1) A car is driven 225 \(\mathrm{km}\) west and then 78 \(\mathrm{km}\) southwest \(\left(45^{\circ}\right) .\) What is the displacement of the car from the point of origin (magnitude and dircction)? Draw a diagram.
7 step solution
Problem 2
A delivery truck travels 28 blocks north, 16 blocks east, and 26 blocks south. What is its final displacement from the origin? Assume the blocks are equal length.
4 step solution
Problem 2
(1) A delivery truck travels 28 blocks north, 16 blocks east, and 26 blocks south. What is its final displacement from the origin? Assume the blocks are equal length.
5 step solution
Problem 3
If \(V_{x}=7.80\) units and \(V_{y}=-6.40\) units, determine the magnitude and direction of \(\overrightarrow{\mathbf{V}}\).
3 step solution
Problem 3
$$ \begin{array}{l}{\text { (1) If } V_{x}=7.80 \text { units and } V_{y}=-6.40 \text { units, determine the }} \\ {\text { magnitude and direction of } \vec{\mathbf{v}} .}\end{array} $$
4 step solution
Problem 4
Graphically determine the resultant of the following three vector displacements: (1) \(24 \mathrm{~m}, 36^{\circ}\) north of east; (2) \(18 \mathrm{~m}\) \(37^{\circ}\) east of north; and (3) \(26 \mathrm{~m}, 33^{\circ}\) west of south.
7 step solution
Problem 4
(II) Graphically determine the resultant of the following three vector displacements: (1) \(24 \mathrm{~m}, 36^{\circ}\) north of east; (2) \(18 \mathrm{~m}\) \(37^{\circ}\) east of north; and (3) \(26 \mathrm{~m}, 33^{\circ}\) west of south.
4 step solution
Problem 5
\(\overrightarrow{\mathbf{V}}\) is a vector 24.8 units in magnitude and points at an angle of \(23.4^{\circ}\) above the negative \(x\) axis. \((a)\) Sketch this vector. (b) Calculate \(V_{x}\) and \(V_{y} \cdot(c)\) Use \(V_{x}\) and \(V_{y}\) to obtain (again) the magnitude and direction of \(\overrightarrow{\mathbf{V}}\). [Note: Part \((c)\) is a good way to check if you've resolved your vector correctly.]
5 step solution
Problem 5
(II) \(\hat{\mathbf{v}}\) is a vector 24.8 units in magnitude and points at an angle of \(23.4^{\circ}\) above the negative \(x\) axis. \((a)\) Sketch this vector. (b) Calculate \(V_{x}\) and \(V_{y}-(c)\) Use \(V_{x}\) and \(V_{y}\) .to obtain (again) the magnitude and direction of \(\vec{\mathbf{v}}\)[Note. Part \((c)\) is a good way to check if you've resolved your vector correctly.
5 step solution
Problem 7
An airplane is traveling \(835 \mathrm{~km} / \mathrm{h}\) in a direction \(41.5^{\circ}\) west of north (Fig. \(3-37)\). ( \(a\) ) Find the components of the velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane traveled after \(2.50 \mathrm{~h} ?\)
6 step solution
Problem 8
$$ \vec{\mathbf{v}}_{1}=-6.0 \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}} \text { and } \vec{\mathbf{v}}_{2}=4.5 \hat{\mathrm{i}}-5.0 \mathrm{j} . $$ mine the magnitude and direction of $$ (a) \vec{\mathbf{v}}_{1},(b) \vec{\mathbf{v}}_{2} $$ $$ (c) \vec{\mathbf{v}}_{1}+\vec{\mathbf{v}}_{2} \text { and }(d) \vec{\mathbf{v}}_{2}-\vec{\mathbf{v}}_{1-} $$
8 step solution
Problem 9
(a) Determine the magnitude and direction of the sum of the three vectors \(\overrightarrow{\mathbf{v}}_{1}=4.0 \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}}, \overrightarrow{\mathbf{v}}_{2}=\hat{\mathbf{i}}+\hat{\mathbf{j}},\) and \(\overrightarrow{\mathbf{v}}_{3}=-2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}} .(b)\) Determine \(\overrightarrow{\mathbf{v}}_{1}-\overrightarrow{\mathbf{v}}_{2}+\overrightarrow{\mathbf{v}}_{3}\)
4 step solution
Problem 9
(II) (a) Determine the magnitude and direction of the sum of the three vectors $$ \vec{\mathbf{v}}_{1}=4.0 \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}} \cdot \vec{\mathbf{v}}_{2}=\hat{\mathbf{i}}+\hat{\mathbf{j}} $$ $$ \vec{\mathbf{v}}_{3}=-2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}} .(b) \text { Determine } \vec{\mathbf{v}}_{1}-\vec{\mathbf{v}}_{2}+\vec{\mathbf{v}}_{3} $$
5 step solution
Problem 15
The summit of a mountain, \(2450 \mathrm{~m}\) above base camp, is measured on a map to be \(4580 \mathrm{~m}\) horizontally from the camp in a direction \(32.4^{\circ}\) west of north. What are the components of the displacement vector from camp to summit? What is its magnitude? Choose the \(x\) axis east, \(y\) axis north, and \(z\) axis up.
9 step solution
Problem 15
(II) The summit of a mountain, 2450 \(\mathrm{m}\) above base camp, is measured on a map to be 4580 \(\mathrm{m}\) horizontally from the camp in a direction \(32.4^{\circ}\) west of north. What are the components of the displacement vector from camp to summit? What is its magnitude? Choose the \(x\) axis cast, \(y\) axis north, and \(z\) axis up.
7 step solution
Problem 16
You are given a vector in the \(x y\) plane that has a magnitude of 90.0 units and a \(y\) component of -55.0 units. (a) What are the two possibilities for its \(x\) component? (b) Assuming the \(x\) component is known to be positive, specify the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the \(-x\) direction.
5 step solution
Problem 16
(III) You are given a vector in the \(x y\) plane that has a magnitude of 90.0 units and a \(y\) component of \(-55.0\) units. (a) What are the two possibilities for its \(x\) component? (b) Assuming the \(x\) component is known to be positive, specify the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points cntirely in the \(-x\) dircction.
8 step solution
Problem 17
The position of a particular particle as a function of time is given by \(\overrightarrow{\mathbf{r}}=\left(9.60 t \hat{\mathbf{i}}+8.85 \hat{\mathbf{j}}-1.00 t^{2} \hat{\mathbf{k}}\right) \mathrm{m} .\) Determine the particle's velocity and acceleration as a function of time.
3 step solution
Problem 17
(I) The position of a particular particle as a function of time is given by $$ \vec{\mathbf{r}}=\left(9.60 t \hat{\mathbf{i}}+8.85 \hat{\mathbf{j}}-1.00 t^{2} \hat{\mathbf{k}}\right) \mathrm{m} $$ Determine the particles velocity and acceleration as a function of time.
3 step solution
Problem 20
A car is moving with speed \(18.0 \mathrm{~m} / \mathrm{s}\) due south at one moment and \(27.5 \mathrm{~m} / \mathrm{s}\) due east \(8.00 \mathrm{~s}\) later. Over this time interval, determine the magnitude and direction of \((a)\) its average velocity, \((b)\) its average acceleration. \((c)\) What is its average speed. [Hint: Can you determine all these from the information given?]
7 step solution
Problem 20
(II) A car is moving with speed 18.0 \(\mathrm{m} / \mathrm{s}\) due south at one moment and 27.5 \(\mathrm{m} / \mathrm{s}\) due east 8.00 \(\mathrm{s}\) later. Over this time interval, determine the magnitude and direction of \((a)\) its $$ \begin{array}{l}{\text { average velocity, }(b) \text { its average acceleration. }(c) \text { What is its }} \\ {\text { average speed. [Hint: Can you determine all these from the }} \\ {\text { information given? }}\end{array} $$
6 step solution
Problem 21
At \(t=0\), a particle starts from rest at \(x=0, y=0\), and moves in the \(x y\) plane with an acceleration \(\overrightarrow{\mathbf{a}}=(4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) Determine \((a)\) the \(x\) and \(y\) components of velocity, \((b)\) the speed of the particle, and \((c)\) the position of the particle, all as a function of time. \((d)\) Evaluate all the above at \(t=2.0 \mathrm{~s}\).
4 step solution
Problem 21
(II) At \(t=0,\) a particle starts from rest at \(x=0, y=0\) , and moves in the \(x y\) planc with an acceleration $$ \begin{array}{l}{\vec{\mathbf{a}}=(4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} . \text { Determine }(a) \text { the } x \text { and } y \text { compo- }} \\ {\text { nents of velocity, }(b) \text { the spced of the particle, and }(c) \text { the }}\end{array} $$ position of the particle, all as a function of time. (d) Eval- uate all the above at \(t=2.0 \mathrm{s}\) .
4 step solution
Problem 22
\((a)\) A skier is accelerating down a \(30.0^{\circ}\) hill at \(1.80 \mathrm{~m} / \mathrm{s}^{2}\) (Fig. \(3-39\) ). What is the vertical component of her acceleration? (b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is \(325 \mathrm{~m} ?\)
4 step solution
Problem 22
(II) \((a)\) A skier is accelerating down a \(30.0^{\circ}\) hill at 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) (Fig. \(39 ) .\) What is the vertical component of her acceleration? (b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is 325 \(\mathrm{m} ?\)
5 step solution
Problem 23
An ant walks on a piece of graph paper straight along the \(x\) axis a distance of \(10.0 \mathrm{~cm}\) in \(2.00 \mathrm{~s}\). It then turns left \(30.0^{\circ}\) and walks in a straight line another \(10.0 \mathrm{~cm}\) in \(1.80 \mathrm{~s}\). Finally, it turns another \(70.0^{\circ}\) to the left and walks another \(10.0 \mathrm{~cm}\) in \(1.55 \mathrm{~s} .\) Determine \((a)\) the \(x\) and \(y\) components of the ant's average velocity, and \((b)\) its magnitude and direction.
6 step solution
Problem 23
(1I) An ant walks on a piece of graph paper straight along the \(x\) axis a distance of 10.0 \(\mathrm{cm}\) in 2.00 \(\mathrm{s}\) . It then turns left \(30.0^{\circ}\) and walks in a straight line another 10.0 \(\mathrm{cm}\) in 1.80 s Finally, it turns another \(70.0^{\circ}\) to the left and walks another 10.0 \(\mathrm{cm}\) in 1.55 \(\mathrm{s}\) . Determine \((a)\) the \(x\) and \(y\) components of the ant's average velocity, and \((b)\) its magnitude and direction.
4 step solution
Problem 24
(II) A particle starts from the origin at \(t=0\) with an initial velocity of \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive \(x\) axis. If the acceleration is \((-3.0 \hat{\mathbf{i}}+4.5 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2},\) determine the velocity and position of the particle at the moment it reaches its maximum \(x\) coordinate.
5 step solution
Problem 25
(II) Suppose the position of an object is given by \(\vec{\mathbf{r}}=\left(3.0 t^{2} \hat{\mathbf{i}}-6.0 t^{3} \hat{\mathbf{j}}\right) \mathrm{m}\) (a) Determine its velocity \(\vec{\mathbf{v}}\) and acceleration \(\vec{a},\) as a function of time. (b) Determine \(\vec{\mathbf{r}}\) and \(\vec{\mathbf{v}}\) at time \(t=2.5 \mathrm{s} .\)
5 step solution
Problem 26
An object, which is at the origin at time \(t=0\), has initial velocity \(\overrightarrow{\mathbf{v}}_{0}=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) and constant acceleration \(\overrightarrow{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) Find the position \(\overrightarrow{\mathbf{r}}\) where the object comes to rest (momentarily).
5 step solution
Problem 26
(II) An object, which is at the origin at time \(t=0,\) has $$ \overline{\mathbf{v}}_{0}=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} $$ $$ \quad \overline{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} . \text { Find the position } \overline{\mathbf{r}} $$ where the object comes to rest (momentarily).
4 step solution
Problem 27
A particle's position as a function of time \(t\) is given by \(\overrightarrow{\mathbf{r}}=\left(5.0 t+6.0 t^{2}\right) \mathrm{m} \hat{\mathbf{i}}+\left(7.0-3.0 t^{3}\right) \mathrm{m} \hat{\mathbf{j}} .\) At \(t=5.0 \mathrm{~s}\) find the magnitude and direction of the particle's displacement vector \(\Delta \overrightarrow{\mathbf{r}}\) relative to the point \(\overrightarrow{\mathbf{r}}_{0}=(0.0 \hat{\mathbf{i}}+7.0 \hat{\mathbf{j}}) \mathrm{m}\)
4 step solution
Problem 27
(II) A particle's position as a function of time \(t\) is given $$ \vec{\mathbf{r}}=\left(5.0 t+6.0 t^{2}\right) \mathrm{m} \mathbf{i}+\left(7.0-3.0 t^{3}\right) \mathrm{m} \mathbf{j} . \text { At } t=5.0 \mathrm{s} $$ $$ \begin{array}{l}{\text { find the magnitude and direction of the particle's displace- }} \\ {\text { ment vector } \Delta \vec{\mathrm{r}} \text { relative to the point } \mathbf{r}_{0}=(0.0 \mathrm{i}+7.0 \mathrm{j}) \mathrm{m}}\end{array} $$
4 step solution
Problem 28
A tiger leaps horizontally from a 7.5 -m-high rock with a speed of \(3.2 \mathrm{~m} / \mathrm{s}\). How far from the base of the rock will she land?
4 step solution
Problem 28
$$ \begin{array}{l}{\text { (1) A tiger leaps horizontally from a } 7.5 \text { -migh rock with a }} \\ {\text { speed of } 3.2 \mathrm{m} / \mathrm{s} \text { . How far from the base of the rock will she }} \\ {\text { land? }}\end{array} $$
4 step solution
Problem 29
A diver running \(2.3 \mathrm{~m} / \mathrm{s}\) dives out horizontally from the edge of a vertical cliff and \(3.0 \mathrm{~s}\) later reaches the water below. How high was the cliff and how far from its base did the diver hit the water?
2 step solution
Problem 29
(1) A diver running 2.3 \(\mathrm{m} / \mathrm{s}\) dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. How high was the cliff and how far from its base did the diver hit the water?
3 step solution
Problem 30
Estimate how much farther a person can jump on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.
6 step solution
Problem 30
(II) Estimate how much farther a person can jump on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.
4 step solution
Problem 31
(II) A fire hose held near the ground shoots water at a speed of 6.5 \(\mathrm{m} / \mathrm{s}\) . At what angle(s) should the nozzle point in order that the water land 2.5 \(\mathrm{m}\) away (Fig. 40\() ?\) Why are there two different angles? Sketch the two trajectories.
7 step solution
Problem 32
A ball is thrown horizontally from the roof of a building \(9.0 \mathrm{~m}\) tall and lands \(9.5 \mathrm{~m}\) from the base. What was the ball's initial speed?
5 step solution
Problem 32
(II) A ball is thrown horizontally from the roof of a building 9.0 \(\mathrm{m}\) tall and lands 9.5 \(\mathrm{m}\) from the base. What was the ball's initial spced?
4 step solution
Problem 33
A football is kicked at ground level with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(38.0^{\circ}\) to the horizontal. How much later does it hit the ground?
2 step solution
Problem 33
(II) A football is kicked at ground level with a spced of 18.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(38.0^{\circ}\) to the horizontal. How much later does it hit the ground?
6 step solution
Problem 34
A ball thrown horizontally at \(23.7 \mathrm{~m} / \mathrm{s}\) from the roof of a building lands \(31.0 \mathrm{~m}\) from the base of the building. How high is the building?
4 step solution
Problem 34
(II) A ball thrown horizontally at 23.7 \(\mathrm{m} / \mathrm{s}\) from the roof of a building lands 31.0 \(\mathrm{m}\) from the base of the building. How high is the building?
3 step solution
Problem 35
A shot-putter throws the shot (mass \(=7.3 \mathrm{~kg}\) ) with an initial speed of \(14.4 \mathrm{~m} / \mathrm{s}\) at a \(34.0^{\circ}\) angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of \(2.10 \mathrm{~m}\) above the ground.
8 step solution
Problem 35
(II) A shot-putter throws the shot (mass \(=7.3 \mathrm{kg}\) ) with an initial spced of 14.4 \(\mathrm{m} / \mathrm{s}\) at a \(34.0^{\circ}\) angle to the borizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.10 \(\mathrm{m}\) above the ground.
3 step solution
Problem 36
Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height if air resistance is neglible.
6 step solution