Question: Find the components of \({v_{tot}}\) along the \(x\)- and \(y\)-axes in Figure 3.55.

Figure: The two velocities \({{\rm{v}}_{\rm{A}}}\) and \({{\rm{v}}_{\rm{B}}}\) add to give a total \({{\rm{v}}_{{\rm{tot}}}}\).

Short Answer

Expert verified

The \(x\) component of \({v_{tot}}\) is \(4.409\;{\rm{m/s}}\) and \(y\) component of \({v_{tot}}\) is \(5.072\;{\rm{m/s}}\).

Step by step solution

01

Understanding resolution of vectors

Resolution of vectors is the process of breaking a vector into two or more vectors along distinct directions, in order to ascertain the magnitude and the direction of vector components that together generate the same impact as a single vector.

Given data:

  • \({v_{tot}} = 6.72\;{\rm{m/s}}\).
02

Calculating horizontal component of total velocity

Figure: the components and the resultant

The angle between\(x\)-axis and\({v_{tot}}\)is

\(\begin{array}{c}\alpha = \left( {26.5^\circ } \right) + \left( {22.5^\circ } \right)\\ = 49^\circ \end{array}\)

The horizontal component of\({v_{tot}}\)is

\({v_{to{t_x}}} = {v_{tot}}\cos \alpha \)

Substituting\(6.72\;{\rm{m/s}}\)for\({v_{tot}}\)and\(49^\circ \)for\(\alpha \), we have

\(\begin{array}{c}{v_{to{t_x}}} = \left( {6.72\;{\rm{m/s}}} \right) \times \cos \left( {49^\circ } \right)\\ = 4.409\;{\rm{m/s}}\end{array}\)

Hence, the \(x\) component of \({v_{tot}}\) is \(4.409\;{\rm{m/s}}\).

03

Calculating vertical component of horizontal velocity

The vertical component of\({v_{tot}}\)is

\({v_{to{t_y}}} = {v_{tot}}\sin \alpha \)

Substituting\(6.72\;{\rm{m/s}}\)for\({v_{tot}}\)and\(49^\circ \)for\(\alpha \), we have

\(\begin{array}{c}{v_{to{t_y}}} = \left( {6.72\;{\rm{m/s}}} \right) \times \sin \left( {49^\circ } \right)\\ = 5.072\;{\rm{m/s}}\end{array}\)

Hence, the \(y\) component of \({v_{tot}}\) is \(5.072\;{\rm{m/s}}\).

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Most popular questions from this chapter

Find the following for path A in Figure,

(a) the total distance traveled, and

(b) the magnitude and direction of the displacement from start to finish.

The various lines represent paths taken by different people walking in a city. All blocks are120 mon a side.

If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle . What other information would he need to get to Sacramento?

(a) Repeat the problem two problems prior, but for the second leg you walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) north of east (which is equivalent to subtracting \({\rm{B}}\) from \({\rm{A}}\) —that is, to finding \({\rm{R'}} = {\rm{A}} - {\rm{B}}\)).

(b) Repeat the problem two problems prior, but now you first walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) south of west and then \(12.0{\rm{ m}}\) in a direction \(20.0^\circ \) east of south (which is equivalent to subtracting \({\rm{A}}\) from \({\rm{B}}\) —that is, to finding \({\rm{R''}} = {\rm{B}} - {\rm{A}} = - {\rm{R'}}\)). Show that this is the case.

Suppose you walk \({\bf{18}}.{\bf{0}}\)m straight west and then \({\rm{25}}{\rm{.0}}\) m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , then this problem asks you to find their sum R = A + B .)

Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

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