Chapter 2, Problem 33QAP

To determine

The equation for $x(t)$ for each interval of constant acceleration motion for the object whose position as a function of time.

Answer to Problem 33QAP

Between $t=0\text{ s}$ and $t=4.5\text{ s}$ ; $x(t)=6t-12$

Between $t=4\text{ s}$ and $t=5\text{ s}$ ; $x(t)=3t$

Between $t=5\text{ s }$ and $t=8\text{ s}$ ; $x(t)=15\text{ m}$

Between $t=8\text{ s}$ and $t=13\text{ s}$ ; $x(t)=1.5t^2-22.5t+99$

Between $t=13\text{ s}$ and $\text{t = 21 s}$ ; $x(t)=5t-5$

Between $t=21\text{ s}$ and $t\text{=24 s}$ ; $x(t)=-10t+310$

Between $t=24\text{ s}$ and $t=25\text{ s}$ ; $x(t)=70\text{ m}$

Explanation of Solution

Introduction:

We can determine the equation for $x(t)$ at each time interval by plotting a graph of position versus time and calculating the slope and the intercepts manually or using a computer program like MS Excel.

t ( s )	x ( m )
0	-12
1	-6
2	0
3	6
4	12
5	15
6	15
7	15
8	15
9	18
10	24
11	33
12	45
13	60
14	65
15	70
16	75
17	80
18	85
19	90
20	95
21	100
22	90
23	80
24	70
25	70

Between $t=0\text{ s}$ and $t=4\text{ s}$

  $\begin{array}{l}x(t)=\frac{12\text{ m}-\text{(}-\text{12 m)}}{4\text{ s}}t-12\text{ m }\\ x(t)=6t-12\end{array}$

Between $t=4\text{ s}$ and $t=5\text{ s}$

  $\begin{array}{l}x(t)=\frac{15\text{ m}-\text{(12 m)}}{\text{1s}}t\\ x(t)=3t\end{array}$

Between $t=5\text{ s }$ and $t=8\text{ s}$

  $x(t)=15\text{ m}$

Between $t=8\text{ s}$ and $t=13\text{ s}$, there is a quadratic function.

Therefore between $t=8\text{ s}$ and $t=13\text{ s}$ ;

  $x(t)=1.5t^2-22.5t+99$

Between $t=13\text{ s}$ and $\text{t = 21 s}$ ;

Therefore between $t=13\text{ s}$ and $\text{t = 21 s}$ ;

  $x(t)=5t-5$

Between $t=21\text{ s}$ and $t\text{=24 s}$

Therefore between $t=21\text{ s}$ and $t\text{=24 s}$ ;

  $x(t)=-10t+310$

Between $t=24\text{ s}$ and $t=25\text{ s}$ ;

  $x(t)=70\text{ m}$

Conclusion:

Between $t=0\text{ s}$ and $t=4.5\text{ s}$ ; $x(t)=6t-12$

Between $t=4\text{ s}$ and $t=5\text{ s}$ ; $x(t)=3t$

Between $t=5\text{ s }$ and $t=8\text{ s}$ ; $x(t)=15\text{ m}$

Between $t=8\text{ s}$ and $t=13\text{ s}$ ; $x(t)=1.5t^2-22.5t+99$

Between $t=13\text{ s}$ and $\text{t = 21 s}$ ; $x(t)=5t-5$

Between $t=21\text{ s}$ and $t\text{=24 s}$ ; $x(t)=-10t+310$

Between $t=24\text{ s}$ and $t=25\text{ s}$ ; $x(t)=70\text{ m}$