Question

1) Anita is running to the right at 5.0m/s , as shown in the figure. Balls 1 and 2 thrown toward her at 10m/s by friends standing on the ground.(Figure 1)

According to Anita, what is the speed of the first ball?

According to Anita, what is the speed of the second ball?

2)In the Soapbox Derby in (Figure 1) , young participants build non-motorized cars with very low-friction wheels. Cars race by rolling down a hill. Assume that the track begins with a 47-ft-long (1 m = 3.28 ft) section tilted 12? below horizontal.

What is the maximum possible acceleration of a car moving down this stretch of track?

If a car starts from rest and undergoes this acceleration for the full l, what is its final speed in m/s?

3) a) Find the x- and y-components of the vector d?  = (2.0km , 29? left of +y-axis).

b)Find the x- and y-components of the vector v?  = (5.0cm/s , ?x-direction).

c)Find the x- and y-components of the vector a?  = (17m/s2 , 45? left of ?y-axis).

4)

Answer 1

Concepts and reason

The main concept used to solve the problem is relative velocity.

Initially, calculate the relative speed of an object with respect to a moving observer by subtracting the speed of the moving observer from the object speed when the object and observer are moving in same direction and by adding the speed of the moving observer to the speed of the object when the object and observer are moving in opposite direction.

Later, calculate the maximum acceleration possible for the car moving down an incline by using the vertical component of the acceleration due to gravity as gravity is the only force acting on the car.

Finally, calculate the final speed of the car by rearranging kinematics equation of motion and substituting the values. The components of the vector can be calculated by using the sine and cosine function. Substitute the values in the component equation and calculate the corresponding components.

Fundamentals

The relative speed of an object with respect to observer moving in same direction is,

${v_{{\rm{rs}}}} = v - {v_0}$

Here, $v$ is the speed of the object, and ${v_{\rm{O}}}$ is the speed of the observer.

The relative speed of an object with respect to observer moving in opposite direction is,

${v_{{\rm{ro}}}} = v + {v_0}$

Here, $v$ is the speed of the object, and ${v_0}$ is the speed of the observer.

The component of the acceleration due to gravity parallel to the incline plane is,

$a = g\sin \theta$

Here, $g$ is the acceleration due to gravity, and $\theta$ is the angle between the horizontal and incline plane.

The kinematics equation of motion for final velocity in terms of acceleration ad distance travelled is,

$v_{\rm{f}}^2 = v_{\rm{i}}^2 + 2al$

Here, ${v_{\rm{f}}}$ is the final velocity of the object, ${v_{\rm{i}}}$ is the initial velocity, $a$ is the acceleration, and $l$ is the distance travelled by the object.

For unit conversion from feet to meter use $1{\rm{ m}} = 3.28{\rm{ ft}}$ .

The component of a vector $\vec p$ along $x$ axis is,

${p_x} = p\cos \theta$

Here, $p$ is the magnitude of the vector $\vec p$ , and $\theta$ is the angle that vector makes in counterclockwise direction from the $+ x$ axis.

The component of a vector $\vec p$ along $y$ axis is,

${p_y} = p\cos \theta$

Here, $p$ is the magnitude of the vector $\vec p$ , and $\theta$ is the angle that vector makes in counterclockwise direction from the $+ x$ axis.

(1.a)

The relative speed of an object with respect to observer moving in same direction is,

${v_{{\rm{rs}}}} = v - {v_0}$

Here, $v$ is the speed of the object, and ${v_0}$ is the speed of the observer.

Substitute $10{\rm{ m/s}}$ for $v$ , and $5{\rm{ m/s}}$ for ${v_0}$ in the above equation and calculate the speed of the first ball according to Anita.

$\begin{array}{c}\\{v_{{\rm{rs}}}} = 10{\rm{ m/s}} - 5{\rm{ m/s}}\\\\ = 5{\rm{ m/s}}\\\end{array}$

(1.b)

The relative speed of an object with respect to observer moving in opposite direction is,

${v_{{\rm{ro}}}} = v + {v_0}$

Here, $v$ is the speed of the object, and ${v_{\rm{O}}}$ is the speed of the observer.

Substitute $10{\rm{ m/s}}$ for $v$ , and $5{\rm{ m/s}}$ for ${v_{\rm{O}}}$ in the above equation and calculate the speed of the second ball according to Anita.

$\begin{array}{c}\\{v_{{\rm{ro}}}} = 10{\rm{ m/s}} + 5{\rm{ m/s}}\\\\ = 15{\rm{ m/s}}\\\end{array}$

(2.a)

The component of the acceleration due to gravity parallel to the incline plane is,

$a = g\sin \theta$

Here, $g$ is the acceleration due to gravity, and $\theta$ is the angle between the horizontal and incline plane.

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , and $12^\circ$ for $\theta$ in the above equation and calculate the maximum possible acceleration of a car moving down the stretch of track.

$\begin{array}{c}\\a = \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 12^\circ \\\\ = 2.04{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

(2.b)

The kinematics equation of motion for final velocity in terms of acceleration ad distance travelled is,

$v_{\rm{f}}^2 = v_{\rm{i}}^2 + 2al$

Here, ${v_{\rm{f}}}$ is the final velocity of the object, ${v_{\rm{i}}}$ is the initial velocity, $a$ is the acceleration, and $x$ is the distance travelled by the object.

Take square root both side of the above equation and solve for final speed of the car.

${v_{\rm{f}}} = \sqrt {v_{\rm{i}}^2 + 2al}$

Substitute $0{\rm{ m/s}}$ for ${v_{\rm{i}}}$ , $2.04{\rm{ m/}}{{\rm{s}}^2}$ for $a$ , and $47{\rm{ ft}}$ for $l$ in the above equation and calculate the final speed of the car.

${v_{\rm{f}}} = \sqrt {{{\left( {0{\rm{ m/s}}} \right)}^2} + 2\left( {2.04{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {47{\rm{ ft}}} \right)}$

Convert ${\rm{ft}}$ to ${\rm{m}}$ by multiplying with $\left( {\frac{{1{\rm{ m}}}}{{3.28{\rm{ ft}}}}} \right)$ .

$\begin{array}{c}\\{v_{\rm{f}}} = \sqrt {{{\left( {0{\rm{ m/s}}} \right)}^2} + 2\left( {2.04{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {47{\rm{ ft}}} \right)\left( {\frac{{1{\rm{ m}}}}{{3.28{\rm{ ft}}}}} \right)} \\\\ = \sqrt {56.46{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}} \\\\ = 7.45{\rm{ m/s}}\\\end{array}$

(3.a)

Use the components equation to solve for $x$ and $y$ component of the vectors.

$\begin{array}{l}\\{p_x} = p\cos \theta \\\\{p_y} = p\sin \theta \\\end{array}$

Substitute $2.0{\rm{ km}}$ for $p$ , $119^\circ$ for $\theta$ , and ${d_x}$ for ${p_x}$ in the equation ${p_x} = p\cos \theta$ and calculate the $x$ component of the vector $d$ .

$\begin{array}{c}\\{d_x} = \left( {2.0{\rm{ km}}} \right)\cos \left( {119^\circ } \right)\\\\ = - 0.97{\rm{ km}}\\\end{array}$

Substitute $2.0{\rm{ km}}$ for $p$ , $119^\circ$ for $\theta$ , and ${d_y}$ for ${p_y}$ in the equation ${p_y} = p\sin \theta$ and calculate the $y$ component of the vector $d$ .

$\begin{array}{c}\\{d_y} = \left( {2.0{\rm{ km}}} \right)\sin \left( {119^\circ } \right)\\\\ = 1.75{\rm{ km}}\\\end{array}$

(3.b)

Use the components equation to solve for $x$ and $y$ component of the vectors.

$\begin{array}{l}\\{p_x} = p\cos \theta \\\\{p_y} = p\sin \theta \\\end{array}$

Substitute $5.0{\rm{ cm/s}}$ for $p$ , $180^\circ$ for $\theta$ , and ${v_x}$ for ${p_x}$ in the equation ${p_x} = p\cos \theta$ and calculate the $x$ component of the vector $v$ .

$\begin{array}{c}\\{v_x} = \left( {5.0{\rm{ cm/s}}} \right)\cos \left( {180^\circ } \right)\\\\ = - 5.0{\rm{ cm/s}}\\\end{array}$

Substitute $5.0{\rm{ cm/s}}$ for $p$ , $180^\circ$ for $\theta$ , and ${v_y}$ for ${p_y}$ in the equation ${p_y} = p\sin \theta$ and calculate the $y$ component of the vector $v$ .

$\begin{array}{c}\\{v_y} = \left( {5.0{\rm{ cm/s}}} \right)\sin \left( {180^\circ } \right)\\\\ = 0{\rm{ cm/s}}\\\end{array}$

(3.c)

Use the components equation to solve for $x$ and $y$ component of the vectors.

$\begin{array}{l}\\{p_x} = p\cos \theta \\\\{p_y} = p\sin \theta \\\end{array}$

Substitute $17{\rm{ m/}}{{\rm{s}}^2}$ for $p$ , $225^\circ$ for $\theta$ , and ${a_x}$ for ${p_x}$ in the equation ${p_x} = p\cos \theta$ and calculate the $x$ component of the vector $a$ .

$\begin{array}{c}\\{a_x} = \left( {{\rm{17 m/}}{{\rm{s}}^2}} \right)\cos \left( {225^\circ } \right)\\\\ = - 12.02{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Substitute $17{\rm{ m/}}{{\rm{s}}^2}$ for $p$ , $225^\circ$ for $\theta$ , and ${a_y}$ for ${p_y}$ in the equation ${p_y} = p\sin \theta$ and calculate the $y$ component of the vector $a$ .

$\begin{array}{c}\\{a_y} = \left( {{\rm{17 m/}}{{\rm{s}}^2}} \right)\sin \left( {225^\circ } \right)\\\\ = - 12.02{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Ans: Part 1.a

According to Anita, the speed of the first ball is $5{\rm{ m/s}}$ .

Part 1.b

According to Anita, the speed of the second ball is $15{\rm{ m/s}}$ .

Part 2.a

The maximum possible acceleration of a car moving down the stretch of track is $2.04{\rm{ m/}}{{\rm{s}}^2}$ .

Part 2.b

The final speed of the car is $7.5{\rm{ m/s}}$ .

Part 3.a

The $x$ and $y$ components of the vector are $- 0.97{\rm{ km}},{\rm{ 1}}{\rm{.75 km}}$ .

Part 3.b

The $x$ and $y$ components of the vector are $- 5.0{\rm{ cm/s}},{\rm{ 0 cm/s}}$ .

Part 3.c

The $x$ and $y$ components of the vector are $- 12.02{\rm{ m/}}{{\rm{s}}^2},{\rm{ }} - 12.02{\rm{ m/}}{{\rm{s}}^2}$ .