All Solutions
Page 71: Section Review
We, at this point, expect the acceleration to be negative, since it acts opposite to motion…
$v_{f}=v_{i}+overline{a}t$
$d_{f}-d_{i}=v_{i}t+displaystyle frac{1}{2}overline{a}t^{2}$
$v_{f}^{2}=v_{i}^{2}+ 2 overline{a}(d_{f}-d_{i})$
—————————–
$v_{f}^{2}=v_{i}^{2}+ 2 overline{a}(d_{f}-d_{i})$
$displaystyle mathrm{a}=frac{v_{f}^{2}-v_{i}^{2}}{2(d_{mathrm{f}}-d_{mathrm{i}})}$=
$=displaystyle frac{0.0mathrm{m}/mathrm{s}-(23mathrm{m}/mathrm{s})^{2}}{(2)(210mathrm{m})}$=
$$
=-1.3mathrm{m}/mathrm{s}^{2}
$$
– The initial velocity is $v_{text{i}}=23 , frac{text{m}}{text{s}}$
– The final velocity is $v_{text{f}}=0$(since the car stops)
– The distance traveled before stopping $d=210text{ m}$
We need to calculate the acceleration(deceleration) of the car
$$
v_{text{f}}^2 = v_{text{i}}^2 + 2 cdot a cdot d
tag{1}
$$
$$
a = frac{v_{text{f}}^2 – v_{text{i}}^2}{2d}
tag{2}
$$
Now inserting the fact that the final velocity is zero we get:
$$
a= frac{- v_{text{i}}^2}{2d}
tag{3}
$$
$$
begin{align*}
a &= frac{-left(23 , frac{text{m}}{text{s}} right)^2}{2 cdot 210text{ m}}\
&= boxed{-1.3 , frac{text{m}}{text{s}^2}}
end{align*}
$$
This is the required acceleration of the car!
v_{f}^{2} = v_{i}^{2} + 2ad_{f}
$$
$v_{f}=v_{i}+overline{a}t$
$d_{f}-d_{i}=v_{i}t+displaystyle frac{1}{2}overline{a}t^{2}$
$$
v_{f}^{2}=v_{i}^{2}+ 2 overline{a}(d_{f}-d_{i})\\
—————————–\$
1. Accelerating, \$d_{i}=0, v_{i}=0, v_{f}=5.0 m/s, t=4.5 s.\\
$$
$a=displaystyle frac{v_{f}}{t_{1}}=frac{5.00 m/s}{4.5 s}=frac{10}{9}m/s^{2}\\$$From the second formula
d_{f1}=displaystyle $frac{1}{2}$cdot($frac{10}{9}$ m/s^{2})cdot(4.5s)^{2}=11.25m
$2. Constant speed$
d_${mathrm{f}2}$=v_{2}t_{2}=(5.0$mathrm{m}$/$mathrm{s}$)(4.5$mathrm{s}$)=22.5$mathrm{m}$
$total distance$=
11.25$mathrm{m}$+22.5$mathrm{m}$=
=34$$mathrm{m}$
$v_{f}=v_{i}+overline{a}t$
$d_{f}-d_{i}=v_{i}t+displaystyle frac{1}{2}overline{a}t^{2}$
$v_{f}^{2}=v_{i}^{2}+ 2 overline{a}(d_{f}-d_{i})$
$v_{f}^{2}=v_{i}^{2}+ 2 overline{a}(d_{f})$
$v_{f}$=$sqrt{(0.0mathrm{m}/mathrm{s})^{2}+2(5.0mathrm{m}/mathrm{s}^{2})(5.0times 10^{2}mathrm{m})}=71mathrm{m}/mathrm{s}$
– The initial velocity is $v_{text{i}}=0$(since the plane starts from rest)
– The distance traveled is $d=500text{ m}$
– The acceleration of the plane $a=5.0 , frac{text{m}}{text{s}^2}$
We need to find the final velocity of the plane.
$$
v_{text{f}}^2 = v_{text{i}}^2 + 2 cdot a cdot d
tag{1}
$$
$$
v_{text{f}} = sqrt{v_{text{i}}^2 + 2 cdot a cdot d}
tag{2}
$$
Now inserting the fact that the initial velocity is zero we get:
$$
v_{text{f}} = sqrt{2 cdot a cdot d}
tag{3}
$$
$$
begin{align*}
v_{text{f}} &= sqrt{2 cdot left(5.0 , frac{text{m}}{text{s}^2} right) cdot 500text{ m}}\
&= boxed{71 , frac{text{m}}{text{s}}}
end{align*}
$$
This is the final velocity of the plane.
v_{f} = v_{i} + at_{f}
$$
v_{f} = 0 + left(5.0 m/s^{2} right)left( 14 sright)
$$
textbf{underline{textit{Solution}}}
$$
$$
d=v_it+dfrac{1}{2}at^2tag{1}
$$
And now knowing the distance we can use the following equation to find the final speed “after 30.0 s” by which the the airplane took off
$$
{v_f}^2 = {v_i}^2 + 2 a d tag{2}
$$
And it is given that the air plane have an initial speed of zero, and the constant of acceleration by which the airplane is accelerating is 3 m/s$^2$ and the duration of
motion is 30 seconds.
enumerate[bfseries (a)]
item Using equation (1) to find the distance the airplane moved during the 30 seconds
begin{align*}
d&= 0 + dfrac{1}{2} times 3 times (30)^2\
&= 0.50 times 3.0 times (30)^2\
&= fbox{$1350 ~ rm{m}$}
end{align*}
textbf{underline{textit{note:}}} the least number of significant figures is 2 significant figures, therefor the final answer is rounded to 2 significant figure, hence the final results should be 1400 m.\
item And knowing the distance the airplane moved during the 30 seconds, we can find the final velocity the airplane reached using equation (2), we calculate the final velocity as follows
begin{align*}
{v_f}^2 &= 0 + 2 times 3.0 times 1350 \
&= 8100 \
intertext{Taking, the square root to find the value of the final velocity, we get}
v_f &= sqrt{8100}\
&= fbox{$90 ~ rm{m/s}$}
end{align*}
item 1400 m.
item 90 m/s .
$v_{f}=v_{i}+overline{a}t$
$d_{f}-d_{i}=v_{i}t+displaystyle frac{1}{2}overline{a}t^{2}$
$v_{f}^{2}=v_{i}^{2}+ 2 overline{a}(d_{f}-d_{i})$
$——————-\\$
(a)\$d_{i}=0
$
($mathrm{b}$)
v_{f}=v_{i}+$overline{a}$t$\\$=0.0$mathrm{m}$/$mathrm{s}$+(3.00$mathrm{m}$/$mathrm{s}$^{2})(30.0$mathrm{s}$)
=90.0$mathrm{m}$/$$mathrm{s}$
