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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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