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At the instant shown, cars A and B are traveling at velocities of 58 m/s and...

At the instant shown, cars A and B&

At the instant shown, cars A and B are traveling at velocities of 58 m/s and 32 m/s , respectively. B is increasing its velocity by 2 m/s2, while A maintains a constant velocity. The radius of curvature at B is?B = 200 m.

Part A

Determine the magnitude of the velocity of B with respect to A.

Part B

Determine the direction angle of the velocity of B with respect to A.

Part C

Determine the magnitude of the acceleration of B with respect to A.

Part D

Determine the direction angle of the acceleration of B with respect to A.

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Concepts and reason

The concepts used to solve this problem are curvilinear motion and relative velocity.

Curvilinear motion:

A particle is said to follow a curvilinear motion if the particle moves on a path other than a straight line or along a curved path.

As in this problem car BB moves on a curved path so it has a curvilinear motion.

Velocity in a curvilinear motion:

The direction of velocity of car BB , moving along a curved path, is always tangential to the path followed by it. So, the direction of velocity changes at every instant of the motion. The magnitude of velocity also changes at every instant because car BB moves with a given acceleration.

Acceleration in curvilinear motion:

Acceleration is defined as rate of change of velocity. But in case of curvilinear motion both the magnitude and direction of velocity vary with time. So, acceleration in this case has two components tangential and normal component of acceleration. The net acceleration is the vector sum of the both the components of acceleration.

Tangential component of acceleration:

Tangential component comes into effect due to change in the magnitude of velocity at every instant. This component always acts tangentially to the path of the particle.

Centripetal acceleration or normal component of acceleration:

Normal component comes into effect due to change in the direction of velocity at every instant. This component always acts towards the center of the curved path of the particle.

Relative velocity:

It is the velocity of one body with respect to the other body, assuming the other body to be at rest. Relative velocity comes handy when the two bodies are in motion and one has to analyze the motion of one of the body with respect to the other.

Vector Quantity:

These are quantities that require magnitude and direction both to describe them completely. A few examples are velocity, force, acceleration and energy.

Vector Resolution

This is a technique to divide a vector into rectangular components. For the purpose of this problem the vector is divided into two components. The component of a vector along the x-axis is called horizontal component and the component of a vector along the y-axis is called vertical component.

Initially, find the velocity vector of car AA and car BB . Then, determine the relative velocity in vector form. Use expression for magnitude of a vector to determine the relative velocity of car BB with respect to car AA . Then, find the acceleration vector of car AA and car BB . Finally, determine the relative acceleration in form of vector. Use expression for magnitude of a vector to determine the relative acceleration of car BB with respect to car AA .

Fundamentals

Representation of vector in Cartesian form:

A vector is represented in Cartesian form as,

A=(xi^+yj^)\vec A = \left( {x{\bf{\hat i}} + y{\bf{\hat j}}} \right)

Here, xx and yy are the components of the vector on positive x axis and positive y axis respectively and i^{\bf{\hat i}} , j^{\bf{\hat j}} are the unit vectors along x axis and y axis respectively.

Magnitude of A\mathop A\limits^ \to is,

A=x2+y2\left| {\vec A} \right| = \sqrt {{x^2} + {y^2}}

The expression for direction of vector A\vec A is written as,

φ=tan1(yx)\varphi = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)

Here, φ\varphi is the direction of A\vec A with respect to positive x axis.

Vector resolution:

For a vector VV making an angle θ\theta with the positive x axis, the vector resolution to obtain the rectangular components is shown as,

Vsino
> Vcoso

The horizontal component along positive x axis (Vx)\left( {{V_x}} \right) and vertical component along positive y axis (Vy)\left( {{V_y}} \right) is,

Vx=Vcosθ{V_x} = V\cos \theta

Vy=Vsinθ{V_y} = V\sin \theta

Sign Convention:

The component along positive direction of x axis is taken as positive and vice versa.

The component along positive direction of y axis is taken as positive and vice versa.

Relative velocity of car BB with respect to car AA is,

vBA=vBvA{\vec v_{BA}} = {\vec v_B} - {\vec v_A}

Here, vBA{\vec v_{BA}} is the relative velocity of car BB with respect to car AA , vA{\vec v_A} is the velocity of car AA and vB{\vec v_B} is the velocity of car BB .

Relative acceleration of car BB with respect to car AA is,

aBA=aBaA{\vec a_{BA}} = {\vec a_B} - {\vec a_A}

Here, aBA{\vec a_{BA}} is the relative acceleration of car BB with respect to car AA , aA{\vec a_A} is the acceleration of car AA and aB{\vec a_B} is the acceleration of car BB .

Expression for Centripetal acceleration is,

(aB)c=vB2R{\left( {{a_B}} \right)_c} = \frac{{{v_B}^2}}{R}

Here, (aB)c{\left( {{a_B}} \right)_c} is the centripetal acceleration of the car BB , vB{v_B} is the speed of car BB and RR is the radius of the curved path.

(A)

Draw the vector diagram of the given situation for both the cars.

Vg cos30°
Vo sin30°
Car A
Car B

Calculate the velocity of car A as,

vA=vAj^{{\bf{\vec v}}_{\bf{A}}} = {v_A}{\rm{ }}{\bf{\hat j}}

Substitute (58m/s)\left( {{\rm{58 m/s}}} \right) for vA{v_A} .

vA=(58m/s)j^{{\bf{\vec v}}_{\bf{A}}} = \left( {{\rm{58 m/s}}} \right){\rm{ }}{\bf{\hat j}}

Calculate the velocity of car B as,

vB=(vBsin30)i^+(vBcos30)j^{{\bf{\vec v}}_B} = - \left( {{v_B}\sin 30^\circ } \right){\rm{ }}{\bf{\hat i}} + \left( {{v_B}\cos 30^\circ } \right){\rm{ }}{\bf{\hat j}}

Substitute (32m/s)\left( {{\rm{32 m/s}}} \right) for vB{v_B} .

vB=(16m/s)i^+(27.7m/s)j^{{\bf{\vec v}}_B} = - \left( {{\rm{16 m/s}}} \right){\rm{ }}{\bf{\hat i}} + \left( {{\rm{27}}{\rm{.7 m/s}}} \right){\rm{ }}{\bf{\hat j}}

Write the expression for relative velocity of car B with respect to car A.

vBA=vBvA{{\bf{\vec v}}_{{\bf{BA}}}} = {{\bf{\vec v}}_{\bf{B}}} - {{\bf{\vec v}}_{\bf{A}}}

Substitute (58m/s)j^\left( {{\rm{58 m/s}}} \right){\rm{ }}{\bf{\hat j}} for vA{{\bf{\vec v}}_{\bf{A}}} and (16m/s)i^+(27.7m/s)j^ - \left( {{\rm{16 m/s}}} \right){\rm{ }}{\bf{\hat i}} + \left( {{\rm{27}}{\rm{.7 m/s}}} \right){\rm{ }}{\bf{\hat j}} for vB{{\bf{\vec v}}_B} in the above expression of the relative speed expression,

vBA=(16m/s)i^+(27.7m/s)j^(58m/s)j^=(16m/s)i^+(30.3m/s)j^=(16i^30.3j^)m/s\begin{array}{c}\\{{{\bf{\vec v}}}_{{\bf{BA}}}} = - \left( {{\rm{16 m/s}}} \right){\rm{ }}{\bf{\hat i}} + \left( {{\rm{27}}{\rm{.7 m/s}}} \right){\rm{ }}{\bf{\hat j}} - \left( {{\rm{58 m/s}}} \right){\rm{ }}{\bf{\hat j}}\\\\ = - \left( {{\rm{16 m/s}}} \right){\rm{ }}{\bf{\hat i}} + \left( { - 30.3{\rm{ m/s}}} \right){\rm{ }}{\bf{\hat j}}\\\\ = \left( { - 16{\rm{ }}{\bf{\hat i}} - 30.3{\rm{ }}{\bf{\hat j}}} \right){\rm{ m/s}}\\\end{array}

Obtain the expression of the magnitude of the velocity vBA{{\bf{\vec v}}_{{\bf{BA}}}} as,

vBA=(vBA)x2+(vBA)y2\left| {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right| = \sqrt {\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_x^2 + \left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_y^2}

Substitute 16m/s - 16{\rm{ m/s}} for (vBA)x{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_x} and 30.3m/s - 30.3\,{\rm{m/s}} for (vBA)y{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_y} .

vBA=(16)2+(30.3)2m/s=34.3m/s\begin{array}{c}\\\left| {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right| = \sqrt {{{\left( { - 16} \right)}^2} + {{\left( { - 30.3} \right)}^2}} {\rm{ m/s}}\\\\ = 34.3{\rm{ m/s}}\\\end{array}

(B)

Obtain the expression for direction for the relative velocity of car BB with respect to x axis as,

φ=tan1((vBA)y(vBA)x)\varphi = {\tan ^{ - 1}}\left( {\frac{{{{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)}_y}}}{{{{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)}_x}}}} \right)

Here, φ\varphi is the direction angle of velocity for the relative velocity of car B, (vBA)x{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_x} and (vBA)y{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_y} are the x and y component of aAB{{\rm{\vec a}}_{{\rm{AB}}}} .

Substitute 16m/s - 16{\rm{ m/s}} for (vBA)x{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_x} and 30.3m/s - 30.3\,{\rm{m/s}} for (vBA)y{\left( {{{{\bf{\vec v}}}_{{\bf{BA}}}}} \right)_y} in the above expression of the direction of velocity vector.

φ=tan1(30.3m/s16m/s)\varphi = {\tan ^{ - 1}}\left( {\frac{{ - 30.3{\rm{ m/s}}}}{{ - 16{\rm{ m/s}}}}} \right)

Since both the x and y components of vBA{{\bf{\vec v}}_{{\bf{BA}}}} are negative the angle φ\varphi must lie in the third quadrant. This implies that,

φ=242.2\varphi = 242.2^\circ

(C)

Draw the vector diagram for acceleration of car B.

(d, cos 30°)
(d, sin 30°)
(a.cos 30°)j
20°
(a sin 30°)

Acceleration of car A is zero as the velocity is constant.

Acceleration of car B is given as follows:

aB=(atsin30+accos30)i^+(atcos30acsin30)j^=(atsin30+vB2Rcos30)i^+(atcos30vB2Rsin30)j^=(2sin30+322200cos30)i^+(2cos30322200sin30)j^=(5.434i^0.828j^)m/s2\begin{array}{c}\\{{{\bf{\vec a}}}_{\bf{B}}} = - \left( {{a_t}\sin 30^\circ + {a_c}\cos 30^\circ } \right){\bf{\hat i}} + \left( {{a_t}\cos 30^\circ - {a_c}\sin 30^\circ } \right){\bf{\hat j}}\\\\ = - \left( {{a_t}\sin 30^\circ + \frac{{{v_B}^2}}{R}\cos 30^\circ } \right){\rm{ }}{\bf{\hat i}} + \left( {{a_t}\cos 30^\circ - \frac{{{v_B}^2}}{R}\sin 30^\circ } \right){\rm{ }}{\bf{\hat j}}\\\\ = - \left( {2\sin 30^\circ + \frac{{{{32}^2}}}{{200}}\cos 30^\circ } \right){\rm{ }}{\bf{\hat i}} + \left( {2\cos 30^\circ - \frac{{{{32}^2}}}{{200}}\sin 30^\circ } \right){\rm{ }}{\bf{\hat j}}\\\\ = \left( { - 5.434{\rm{ }}{\bf{\hat i}} - 0.828{\rm{ }}{\bf{\hat j}}} \right){\rm{ m/}}{{\rm{s}}^2}\\\end{array}

Obtain the expression for relative acceleration of car B to that of car A as,

aBA=aBaA{{\bf{\vec a}}_{{\bf{BA}}}} = {{\bf{\vec a}}_{\bf{B}}} - {{\bf{\vec a}}_{\bf{A}}}

Substitute (5.434i^0.828j^)m/s2\left( { - 5.434{\rm{ }}{\bf{\hat i}} - 0.828{\rm{ }}{\bf{\hat j}}} \right){\rm{ m/}}{{\rm{s}}^2} for aB{{\bf{\vec a}}_{\bf{B}}} and 00 for aA{{\bf{\vec a}}_{\bf{A}}} .

aBA=(5.434i^0.828j^)m/s20m/s2=(5.434i^0.828j^)m/s2\begin{array}{c}\\{{{\bf{\vec a}}}_{{\bf{BA}}}} = \left( { - 5.434{\rm{ }}{\bf{\hat i}} - 0.828{\rm{ }}{\bf{\hat j}}} \right){\rm{ m/}}{{\rm{s}}^2} - 0{\rm{ m/}}{{\rm{s}}^2}\\\\ = \left( { - 5.434{\rm{ }}{\bf{\hat i}} - 0.828{\rm{ }}{\bf{\hat j}}} \right){\rm{ m/}}{{\rm{s}}^2}\\\end{array}

From the magnitude expression the expression of the magnitude of the relative acceleration of car B is expresses as follows,

aBA=(5.434)2+(0.828)2m/s2=5.497m/s25.5m/s2\begin{array}{c}\\\left| {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right| = \sqrt {{{\left( { - 5.434} \right)}^2} + {{\left( { - 0.828} \right)}^2}} {\rm{ m/}}{{\rm{s}}^2}\\\\ = 5.497{\rm{ m/}}{{\rm{s}}^2}\\\\ \simeq 5.5{\rm{ m/}}{{\rm{s}}^2}\\\end{array}

(D)

Write the expression of the direction angle for the relative acceleration of car BB with respect to x axis as,

α=tan1((aBA)y(aBA)x)\alpha = {\tan ^{ - 1}}\left( {\frac{{{{\left( {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right)}_y}}}{{{{\left( {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right)}_x}}}} \right)

Here, α\alpha is the direction acceleration for the relative acceleration of car B, (aBA)x{\left( {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right)_{\bf{x}}} and (aBA)y{\left( {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right)_{\bf{y}}} are the x and y component of aAB{{\bf{\vec a}}_{{\bf{AB}}}} .

Substitute 0.828m/s2 - 0.828{\rm{ m/}}{{\rm{s}}^2} for (aBA)y{\left( {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right)_{\bf{y}}} and 5.434m/s2 - 5.434{\rm{ m/}}{{\rm{s}}^2} for (aBA)x{\left( {{{{\bf{\vec a}}}_{{\bf{BA}}}}} \right)_{\bf{x}}} .

α=tan1(0.828m/s25.434m/s2)\alpha = {\tan ^{ - 1}}\left( {\frac{{ - 0.828{\rm{ m/}}{{\rm{s}}^2}}}{{ - 5.434{\rm{ m/}}{{\rm{s}}^2}}}} \right)

Since both the x and y components of aBA{{\bf{\vec a}}_{{\bf{BA}}}} are negative the angle α\alpha must lie in the third quadrant. This implies that,

α=188.67\alpha = 188.67^\circ

Ans: Part A

The magnitude of velocity B with respect to A is equal to 34.3m/s34.3{\rm{ m/s}} .

Part B

The direction angle of the velocity of BB with respect to AA is equal to 242.2242.2^\circ .

Part C

The magnitude of acceleration of BB with respect to AA is equal to 5.5m/s25.5{\rm{ m/}}{{\rm{s}}^2} .

Part D

The direction angle of the acceleration component of BB with respect to AA is equal to 188.67188.67^\circ .

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