Answer
See the detailed answer below.
Work Step by Step
We have here to draw five graphs, from the given velocity-time graph, which are
a) The acceleration-time graph.
(We know that the acceleration is given by $a_x=dv_x/dt$ which is the slope of the given graph).
b) The position-time graph.
(We also know that the position is given by $x=\int v_xdt$ which is the area under the velocity-time graph).
c) The kinetic energy-time graph.
( The kinetic energy is given by $K=\frac{1}{2}mv_x^2$).
d) The force-time graph.
( The force is given by $F=ma_x$).
g) The force-position graph.
The author told us to draw them by calculating and plotting the numerical values of given times, as we see in the table below.
\begin{array}{|c|c|c|}
\hline
t({\rm s})& v_x({\rm m/s})& a_x({\rm m/s^2})&K({\rm J})&x({\rm m})&F({\rm N})\\
\hline
0 & 0 & 10& 0 &0 &5\\
\hline
1 & 10 & 10 &25&5 &5\\
\hline
2 & 20 & 10,{\rm or}\;-10&100&20 &5,{\rm or} -5 \\
\hline
3 & 10 & -10,{\rm or}\;0&25&35 &-5,{\rm or}\;0\\
\hline
4 & 10 & 0&25&45 & 0\\
\hline
\end{array}
---
e) Now we need to find the impulse delivered to the object during the two-time intervals of $t=0\rm\;s\rightarrow 2\;s$ and $t=2\rm\;s\rightarrow 4\;s$.
We can use graph (D) below.
$$J=\int_0^tF_xdt$$
Thus,
$$J_1=\int_0^{2\rm s}5dt=5t\bigg|_0^2=10-0$$
$$J_1=\color{red}{\bf 10}\;\rm N.s $$
And,
$$J_2=\int_{2\rm s}^{3\rm s}F_xdt+\int_{3\rm s}^{4\rm s}F_xdt$$
$$J_2=\int_{2\rm s}^{3\rm s}-5dt+\int_{3\rm s}^{4\rm s}0dt$$
$$J_2=\int_{2\rm s}^{3\rm s}-5dt+0=-5t\bigg|_2^3=-5(4-2)$$
$$J_2=\color{red}{\bf -5}\;\rm N.s $$
-----------------------------------------
f) The author told us to use the impulse-momentum theorem to find the velocity of the particle.
$$J=\Delta p$$
$$J=mv_f-mv_i$$
Solving for $v_f$;
$$v_f=\dfrac{J}{m}+v_i$$
Thus, at $t=2$s,
$$v_f=\dfrac{J_1}{m}+0$$
where $v_i=v_0=0$ m/s,
Plugging the known;
$$v_f=\dfrac{10}{0.5}=\color{red}{\bf 20}\;\rm m/s$$
And, at $t=4$s,
$$v_f=\dfrac{J_2}{m}+v_{2}$$
where $v_i=v_2=20$ m/s,
Plugging the known;
$$v_f=\dfrac{-5}{0.5}+20=\color{red}{\bf 10}\;\rm m/s$$
Therefore, yes, the results agree with the velocity in the graph.
-----------------------------------------
h) We know that the work is given by
$$W=Fd$$
Thus, the work done from 0 s to 2 s is given by
$$W_1=F_1d_1=5\times 20$$
$$W_1=\color{red}{\bf100}\;\rm J$$
And, the work done from 2 s to 4 s is given by
$$W_2=F_2d_2=(-5\times [35-20])+(0\times [45-35]) $$
$$W_2=\color{red}{\bf -75}\;\rm J$$
-----------------------------------------
i) We need to use the work-kinetic energy theorem to find the velocity at $t=2$ s and $t=4$ s.
$$W=\Delta K =K_f-K_i$$
$$W =\frac{1}{2}m v_f^2-\frac{1}{2}m v_i^2 $$
Thus,
$$\dfrac{2W + m v_i^2 }{m}= v_f^2 $$
$$ v_f=\sqrt{ \dfrac{2W + m v_i^2 }{m}}$$
- In the first stage, $v_i=0$ and $W_1=100$ J.
Thus,
$$ v_f=\sqrt{ \dfrac{(2\times 100) + 0.5(0)^2 }{0.5}}=\color{red}{\bf 20}\;\rm m/s$$
- In the second stage, $v_i=20$ and $W_2=-75$ J.
Thus,
$$ v_f=\sqrt{ \dfrac{(2\times -75) + 0.5(20)^2 }{0.5}}=\color{red}{\bf 10}\;\rm m/s$$
Therefore, yes, the results agree with the velocity in the graph.
