Chapter 3, Problem 43QAP

To determine

(i) The x component of the velocity versus time

(ii) The y component of the velocity versus time

(iii) The y component of the acceleration versus time.

Explanation of Solution

Given info:

Magnitude of the initial velocity

  $v_i=20\text{ m/s}$

Angle made by v_i with x- axis

  ${\theta }_i=60°\text{ below the }-x\text{ axis}$

Magnitude of the final velocity

  $v_f=20\text{ m/s}$

Angle made by v_f with x axis

  ${\theta }_f=30°\text{ above the }+x\text{ axis}$

Time interval during which the velocity changes

  $\Delta t=10\text{ s}$

Formula used:

The components of the velocity vectors along the x and the y directions are given by,

  $\begin{array}{l}{v}_{ix}=v_i\cos {\theta }_i\\ v_{iy}=v_i\sin {\theta }_i\\ v_{fx}=v_f\cos {\theta }_f\\ v_{fy}=v_f\sin {\theta }_f\end{array}$

The change in velocity along the x and the y directions is given by,

  $\begin{array}{l}\Delta v_x=v_{fx}-v_{ix}\\ \Delta v_y=v_{fy}-v_{iy}\end{array}$

The components of the acceleration along the x and y directions are given by,

  $\begin{array}{c}{a}_x=\frac{\Delta v_x}{\Delta t}\\ a_y=\frac{\Delta v_y}{\Delta t}\end{array}$

The equation to determine the variation of $v_x$ with time is given by,

  $v_x=v_{ix}+a_{x}t$

The equation to determine the variation of $v_y$ with time is given by,

  $v_y=v_{iy}+a_{y}t$

Calculation:

Resolve the vectors v_i and v_f along the x and the y axes. This is shown in the diagram below.

The components of the velocity vector v_i are directed along +x and-y axes, hence the y component is assigned a negative sign. Calculate the magnitude of the components substituting the values of the variables in the equation,

  $\begin{array}{c}{v}_{ix}=v_i\cos {\theta }_i\\ =\left(20\text{ m/s}\right)\left(\cos 60°\right)\\ =10.0\text{ m/s}\\ v_{iy}=v_i\sin {\theta }_i\\ =-\left(20\text{ m/s}\right)\left(\sin 60°\right)\\ =-17.32\text{ m/s}\end{array}$

The components of the final velocity vector v_f are directed along the +x and +y directions, hence the components are positive. Calculate the magnitude of the components of the vector v_f using the values of the variables.

  $\begin{array}{c}{v}_{fx}=v_f\cos {\theta }_f\\ =\left(20\text{ m/s}\right)\left(\cos 30°\right)\\ =17.32\text{ m/s}\\ v_{fy}=v_f\sin {\theta }_f\\ \left(20\text{ m/s}\right)\left(\sin 30°\right)\\ =10.0\text{ m/s}\end{array}$

Calculate the change in velocity along the x and the y directions.

  $\begin{array}{c}\Delta v_x=v_{fx}-v_{ix}\\ =\left(17.32\text{ m/s}\right)-\left(10.0\text{ m/s}\right)\\ =7.32\text{ m/s}\\ \Delta v_y=v_{fy}-v_{iy}\\ =\left(10.0\text{ m/s}\right)-\left(-17.32\text{ m/s}\right)\\ =27.32\text{ m/s}\end{array}$

Calculate the components of the acceleration along the x and the y directions.

  $\begin{array}{c}{a}_x=\frac{\Delta v_x}{\Delta t}\\ =\frac{7.32\text{ m/s}}{10\text{ s}}\\ =0.732{\text{ m/s}}^2\\ a_y=\frac{\Delta v_y}{\Delta t}\\ =\frac{27.32\text{ m/s}}{10\text{ s}}\\ =2.732{\text{ m/s}}^2\end{array}$

Using the values of v_ix and a_x in the equation $v_x=v_{ix}+a_{x}t$, write the equation to be entered in the spread sheet.

  $v_x=\left(10.0\text{ m/s}\right)+\left(0.732{\text{ m/s}}^2\right)t$

Plot a graph showing the variation of v_x with time.

t in s	vxin m/s
0	10
1	10.732
2	11.464
3	12.196
4	12.928
5	13.66
6	14.392
7	15.124
8	15.856
9	16.588
10	17.32

Using the values of v_iy and a_y in the equation $v_y=v_{iy}+a_{y}t$, write the equation to be entered in the spread sheet.

  $v_y=\left(-17.32\text{ m/s}\right)+\left(2.732{\text{ m/s}}^2\right)t$

Use the equation in a spread sheet and plot the graph showing the variation of v_y with time t.

X-Values	Y-Values
0	-17.32
1	-14.588
2	-11.856
3	-9.124
4	-6.392
5	-3.66
6	-0.928
7	1.804
8	4.536
9	7.268
10	10

The y component of the acceleration a_y is independent of time, since the object's acceleration remains constant during the time interval.

Plot a graph showing the variation of a_y with time t.

t in s	ayin m/s2
0	2.732
1	2.732
2	2.732
3	2.732
4	2.732
5	2.732
6	2.732
7	2.732
8	2.732
9	2.732
10	2.732

Conclusion:

The graph showing the variation of v_x with time is linear and the value increases from 10 m/s to 17.32 m/s at the end of 10 s.

The graph showing the variation of v_y with time is also linear and its value increases from -17.32 m/s to reach a value of 10 m/s at the end of 10 s.

The graph showing the variation of a_y with time is a straight line parallel to the time axis, showing that its value is invariant with time.

Thus all the results are consistent with an object moving with a constant acceleration during the time interval of 10s.