Chapter 2, Problem 20P

(a)

To determine

To find: The velocity of the car as a function of time.

Explanation of Solution

Given:

$t\left(\text{s}\right)$	$x\left(\text{m}\right)$
0	0
0.25	0.11
0.50	0.46
0.75	1.06
1.00	1.94
1.50	4.62
2.00	8.55
2.50	13.79
3.00	20.36
3.50	28.31
4.00	37.65
4.50	48.37
5.00	60.30
5.50	73.26
6.00	87.16

Formula used:

  $v=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}$

Where,

  $v=$ Velocity of the car.

  $\Delta x=$ Displacement of the car.

  $\Delta t=$ Time interval.

Calculation:

Velocity of the car between interval $0\text{ s}-0.25\text{ s}$ :

  $\begin{array}{c}{v}_1=\frac{0.11-0.00}{0.25-0}\\ =0.44\text{ m/s}\end{array}$

The car has this velocity at the midpoint of the interval. i.e. $\left(\frac{0.0+0.25}{2}=1.25\text{ s}\right)$

Velocity of the car between interval $0.25\text{ s}-0.5\text{ s}$ :

  $\begin{array}{c}{v}_2=\frac{0.46-0.11}{0.5-0.25}\\ =1.4\text{ m/s}\end{array}$

  $t\text{'}\left(\text{s}\right)$

The car has this velocity at the midpoint of the interval, $0.375\text{ s}$

$t\left(\text{s}\right)$	$x\left(\text{m}\right)$	$t\text{'}\left(\text{s}\right)$	$V\left(\text{m/s}\right)$
0	0	0	0.00
0.25	0.11	0.125	0.44
0.50	0.46	0.375	1.40
0.75	1.06	0.625	2.40
1.00	1.94	0.875	3.52
1.50	4.62	1.25	5.36
2.00	8.55	1.75	7.86
2.50	13.79	2.25	10.48
3.00	20.36	2.75	13.14
3.50	28.31	3.25	15.90
4.00	37.65	3.73	18.68
4.50	48.37	4.25	21.44
5.00	60.30	4.75	23.86
5.50	73.26	5.25	25.92
6.00	87.16	5.72	27.80

The plot of velocity vs time

Conclusion:

Hence, the velocity in m/s as a function of time is:

  $0.00,0.44,1.40,2.40,3.52,5.36,7.86,10.48,13.14,15.90,18.68,21.44,23.86,25.92,27.80$

(b)

To determine

To find: The acceleration of the car as a function of time.

Explanation of Solution

Given:

$t\text{'}\left(\text{s}\right)$	$V\left(\text{m/s}\right)$
0	0.00
0.125	0.44
0.375	1.40
0.625	2.40
0.875	3.52
1.25	5.36
1.75	7.86
2.25	10.48
2.75	13.14
3.25	15.90
3.73	18.68
4.25	21.44
4.75	23.86
5.25	25.92
5.72	27.80

Formula used:

  $a=\frac{\Delta V}{\Delta t}=\frac{V_2-V_1}{t_2-t_1}$

Where,

  $a=$ Acceleration of the car.

  $\Delta V=$ change in the velocity.

  $\Delta t=$ Time interval.

Calculation:

Acceleration of the car between $0\text{ s}-0.125\text{ s}$ :

  $\begin{array}{c}{a}_1=\frac{0.44-0.00}{0.125-0.00}\\ =3.52{\text{ m/s}}^2\end{array}$

The car has this acceleration at the midpoint of time interval. i.e. $\left(\frac{0.0+0.125}{2}=0.063\text{ s}\right)$

Therefore, the table of acceleration as the function of time:

$t\text{'}\left(\text{s}\right)$	$V\left(\text{m/s}\right)$	$t\text{'}\text{'}\left(\text{s}\right)$	$a\left({\text{m/s}}^2\right)$
0	0.00	0.00	0
0.125	0.44	0.063	3.52
0.375	1.40	0.25	3.84
0.625	2.40	0.50	4.00
0.875	3.52	0.75	4.48
1.25	5.36	1.06	4.91
1.75	7.86	1.50	5.00
2.25	10.48	2.00	5.24
2.75	13.14	2.50	5.32
3.25	15.90	3.00	5.52
3.73	18.68	3.50	5.56
4.25	21.44	4.00	5.52
4.75	23.86	4.50	4.84
5.25	25.92	5.00	4.12
5.72	27.80	5.50	3.76

The plot of acceleration vs time:

   Conclusion:

Hence, the acceleration in ${\text{m/s}}^2$ as a function of time is:

  $0,3.52,3.84,4.00,4.48,4.91,5.00,5.24,5.52,5.56,5.52,4.84,4.12,3.76$