Chapter 2 - Motion Along a Straight Line - Problems - Page 37: 80e

The required graphs are as follows:

We have given distance $x=160$ $m$, final velocity $v=50$ $m/s$ and initial velocity $u=30$ $m/s$. Using the equation of motion $2ax=v^2-u^2$. We get, $2\times a\times 160=50^2-30^2 \implies a=\dfrac{50^2-30^2}{2\times160}\implies a=5$ $m/s^2$. Thus, the constant acceleration of the train is $5$ $m/s^2$. Since, initially train was at rest, initial velocity $u=0$ $m/s$. We have given final velocity $v=50$ $m/s$. Now, using the equation of motion $v=u+at$. We get, $50=0+5\times t \implies t=\dfrac{50}{5}\implies t=10$ $s$. Thus, the total time interval is from $0$ $s$ to $10$ $s$. Substitute $u=0$ $m/s$ and $a=5$ $m/s^2$ in the equation of motion $x=ut+\dfrac{1}{2}at^2$. We get, $x=0\times t+\dfrac{1}{2}\times 5t^2$. $\implies x=2.5t^2$ Thus, the graph of x versus t is parabolic. Take $t=0$ $s$. We get, $x=0$ $m$. Take $t=10$ $s$. We get, $x=250$ $m$. Thus, the endpoints are $(0,0)$ and $(10,250)$. Now, plot the parabolic curve from point $(0,0)$ to point $(10,250)$. Substitute $u=0$ $m/s$ and $a=5$ $m/s^2$ in the equation of motion $v=u+at$. We get, $v=0+5t$. $\implies v=5t$ Thus, the graph of v versus t is linear. Take $t=0$ $s$. We get, $v=0$ $m/s$. Take $t=10$ $s$. We get, $v=50$ $m/s$. Thus, the endpoints are $(0,0)$ and $(10,50)$. Now, plot the straight line from point $(0,0)$ to point $(10,50)$.