第3章直線上の運動

a. $A={\text{m/s}}^{2}B={\text{m/s}}^{5\text{/}2}$;
b. $\begin{array}{}\\ \\ v(t)={\displaystyle \int a(t)dt+C_1}={\displaystyle \int \left(A-B{t}^{1\text{/}2}\right)dt+C_1}=At-\frac{2}{3}B{t}^{3\text{/}2}+C_1\hfill \\ v(0)=0=C_1\text{so}v(t_0)=A{t}_0-\frac{2}{3}B{t}_0^{\text{3/2}}\hfill \end{array}$;
c. $\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+C_2={\displaystyle \int \left(At-\frac{2}{3}B{t}^{3\text{/}2}\right)dt+C_2}}=\frac{1}{2}A{t}^2-\frac{4}{15}B{t}^{5\text{/}2}+C_2\hfill \\ x(0)=0=C_2\text{so}x(t_0)=\frac{1}{2}A{t}_0^2-\frac{4}{15}B{t}_0^{\text{5/2}}\hfill \end{array}$

a. $\begin{array}{}\\ \\ a(t)=3.2{\text{m/s}}^{2}t\le 5.0\text{s}\hfill \\ a(t)=1.5{\text{m/s}}^{2}5.0\text{s}\le t\le 11.0\text{s}\hfill \\ a(t)=0{\text{m/s}}^{2}t>11.0\text{s}\hfill \end{array}$;
b. $\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+C_2={\displaystyle \int 3.2tdt+C_2=1.6t^2+C_2}}\hfill \\ t\le 5.0\text{s}\hfill \\ x(0)=0\Rightarrow C_2=0\text{therefore,}x(2.0\text{s})=6.4\text{m}\hfill \\ x(t)={\displaystyle \int v(t)dt+C_2={\displaystyle \int \left[16.0-1.5\left(t-5.0\right)\right]}dt+C_2}=16t-1.5\left(\frac{t^2}{2}-5.0t\right)+C_2\hfill \\ 5.0\le t\le 11.0\text{s}\hfill \\ x(5\text{s})=1.6{(5.0)}^2=40\text{m}=16(5.0\text{s})-1.5\left(\frac{5^2}{2}-5.0\left(5.0\right)\right)+C_2\hfill \\ 40=98.75+C_2\Rightarrow C_2=\mathrm{-58.75}\hfill \\ x(7.0\text{s})=16(7.0)-1.5\left(\frac{7^2}{2}-5.0\left(7\right)\right)-58.75=69\text{m}\hfill \\ x(t)={\displaystyle \int 7.0dt+C_2=7t+C_2}\hfill \\ t\ge 11.0\text{s}\hfill \\ x(11.0\text{s})=16(11)-1.5\left(\frac{{11}^2}{2}-5.0\left(11\right)\right)-58.75=109=7(11.0\text{s})+C_2\Rightarrow C_2=32\text{m}\hfill \\ x(t)=7t+32\text{m}\hfill \\ x\ge 11.0\text{s}\Rightarrow x(12.0\text{s})=7(12)+32=116\text{m}\hfill \end{array}$

Take west to be the positive direction.
1st plane: $\stackrel{\text{–}}{\nu }=600\text{km/h}$
2nd plane $\stackrel{\text{–}}{\nu }=667.0\text{km/h}$

a.
$v=8.7\times {10}^5\text{m/s}$;
b. $t=7.8\times {10}^{-8}\text{s}$

$1\text{km}=v_0(80.0\text{s})+\frac{1}{2}a{(80.0)}^2$; $2\text{km}=v_0(200.0)+\frac{1}{2}a{(200.0)}^2$ solve simultaneously to get $a=-\frac{0.1}{2400.0}{\text{km/s}}^2$ and $v_0=0.014167\text{km/s}$, which is $51.0\text{km/h}$. Velocity at the end of the trip is $v=21.0\text{km/h}$.

Equation for the speeding car: This car has a constant velocity, which is the average velocity, and is not accelerating, so use the equation for displacement with $x_0=0$:$x=x_0+\stackrel{\text{–}}{v}t=\stackrel{\text{–}}{v}t$; Equation for the police car: This car is accelerating, so use the equation for displacement with $x_0=0$ and $v_0=0$, since the police car starts from rest: $x=x_0+v_{0}t+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2$; Now we have an equation of motion for each car with a common parameter, which can be eliminated to find the solution. In this case, we solve for $t$. Step 1, eliminating $x$: $x=\stackrel{\text{–}}{v}t=\frac{1}{2}a{t}^2$; Step 2, solving for $t$: $t=\frac{2\stackrel{\text{–}}{v}}{a}$. The speeding car has a constant velocity of 40 m/s, which is its average velocity. The acceleration of the police car is 4 m/s². Evaluating t, the time for the police car to reach the speeding car, we have $t=\frac{2\stackrel{\text{–}}{v}}{a}=\frac{2(40)}{4}=20\text{s}$.

At this acceleration she comes to a full stop in $t=\frac{-v_0}{a}=\frac{8}{0.5}=16\text{s}$, but the distance covered is $x=8\text{m/s(16}\text{s)}-\frac{1}{2}(0.5){(16\text{s})}^2=64\text{m}$, which is less than the distance she is away from the finish line, so she never finishes the race.

$v_0=7.9\text{m/s}$ velocity at the bottom of the window.
$v=7.9\text{m/s}$
$v_0=14.5\text{m/s}$

a. $v=5.42\text{m/s}$;
b. $v=4.64\text{m/s}$;
c. $a=2874.28{\text{m/s}}^2$;
d. $(x-x_0)=5.11\times {10}^{-3}\text{m}$

Consider the players fall from rest at the height 1.0 m and 0.3 m.
0.9 s
0.5 s

a. $t=6.37\text{s}$ taking the positive root;
b. $v=59.5\text{m/s}$

a. $y=4.9\text{m}$;
b. $v=38.3\text{m/s}$;
c. $\mathrm{-33.3}\text{m}$

$h=\frac{1}{2}g{t}^2$, h = total height and time to drop to ground
$\frac{2}{3}h=\frac{1}{2}g{(t-1)}^2$ in t – 1 seconds it drops 2/3h
$\frac{2}{3}\left(\frac{1}{2}g{t}^2\right)=\frac{1}{2}g{(t-1)}^2$ or $\frac{t^2}{3}=\frac{1}{2}{(t-1)}^2$
$0=t^2-6t+3$ $t=\frac{6\pm \sqrt{6^2-4·3}}{2}=3\pm \frac{\sqrt{24}}{2}$
t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s $h=\frac{1}{2}g{t}^2=97.0$ m $=\frac{2}{3}(145.5)$