Question

The $v - s$ graph describing the motion of a motorcycle is shown in the figure. The time needed for the motorcycle to reach the position $s = 120\;m$ [Given that $\ln \;5 = 1.6$ ] is $2\;x\;\sec$ , find $x$ ?

Answer 1

Here we find the relation between the velocity $v$ and the displacement $s$ for the two curves separately. For the first half of the journey till $s = 60\;m$ we will formulate the equation of the curve as $v = \dfrac{s}{5} + 3$ . For the second half of the journey from $s = 60\;m$ to $s = 120{\mkern 1mu} m$ , we see that $v = 15\;m/s$ . In both these equations, we will substitute $v = \dfrac{{ds}}{{dt}}$ and solve the integrations with appropriate limits.

Formula used:
$v = \dfrac{{ds}}{{dt}}$

Complete Step by step solution:
From the graph, we form the relationship between the velocity $v$ and the displacement $s$ for the first half of the journey, i.e. from $s = 0\;m$ to $s = 60\;m$ .
The slope of the graph can be determined by dividing the $Y$ axis coordinates with the $X$ axis coordinates. Thus we get the slope as
$\dfrac{{15 - 3}}{{60 - 0}} = \dfrac{1}{5}$
Now considering the general $Y$ axis coordinate as $v$ and the $X$ axis coordinate as $s$ the equation of the line in slope point form will be
$\dfrac{{v - 3}}{{s - 0}} = \dfrac{1}{5}$
$\Rightarrow v = \dfrac{{s + 15}}{5}$
We know that
$v = \dfrac{{ds}}{{dt}}$
Substituting the value of $v$ from the above equation, we get
$\dfrac{{ds}}{{dt}} = \dfrac{{s + 15}}{5}$
$\Rightarrow \dfrac{{ds}}{{s + 15}} = \dfrac{{dt}}{5}$
Now integrating both sides of the equation, we get,
$\Rightarrow \int {\dfrac{{ds}}{{s + 15}}} = \int {\dfrac{{dt}}{5}}$
Putting the limits of integration from $s = 0\;m$ to $s = 60\;m$ , we get,
$\Rightarrow \int\limits_{s = 0}^{s = 60} {\dfrac{{ds}}{{s + 15}}} = \int\limits_{t = 0}^{t = t} {\dfrac{{dt}}{5}}$
$\Rightarrow \left[ {\ln \;(s + 15)} \right]_0^{60} = \left[ {\dfrac{t}{5}} \right]_0^t$
Putting the values of the limits of the integration, we get,
$\Rightarrow \left[ {\ln \;(60 + 15) - \ln \;(0 + 15)} \right] = \left[ {\dfrac{t}{5} - \dfrac{0}{5}} \right]$
$\Rightarrow \left[ {\ln \;(75) - \ln \;(15)} \right] = \dfrac{t}{5}$
On further simplifying, we get,
$\Rightarrow \dfrac{t}{5} = \ln \left( {\dfrac{{75}}{{15}}} \right)$
$\Rightarrow \dfrac{t}{5} = \ln \left( 5 \right)$
$\Rightarrow t = 5 \times \ln \left( 5 \right)$
It is given in the question that $\ln \;5 = 1.6$ . Substituting this value in the above equation, we get,
$\Rightarrow t = 5 \times 1.6$
$\Rightarrow t = 8\;\sec$
This is the time taken for the motorcycle to reach $s = 60\;m$ .
Now for the next half of the journey, i.e. from $s = 60\;m$ to $s = 120\;m$ , we can repeat the above process.
We see that the velocity for this half of the journey is constant, i.e. $v = 15\;m/s$ .
Using this uniform velocity, and the relation $v = \dfrac{{ds}}{{dt}}$ , we get,
$v = 15\;m/s = \dfrac{{ds}}{{dt}}$
$\Rightarrow \dfrac{{ds}}{{dt}} = 15$
$\Rightarrow ds = 15dt$
Integrating both sides, we get,
$\Rightarrow \int {ds} = \int {15dt}$
We now have to substitute the limits carefully. The displacement $s$ will vary from $s = 60\;m$ to $s = 120\;m$ . The time taken $t$ will vary from $t = 8\;\sec$ to $t = t$ , such that the value of the time $t$ will give the actual time taken by the motorcycle to travel the entire distance of $120\;m$ .
$\Rightarrow \int\limits_{s = 60}^{s = 120} {ds} = \int\limits_{t = 8}^{t = t} {15dt}$
$\Rightarrow \left[ s \right]_{60}^{120} = \left[ {15t} \right]_8^t$
Now calculating the limits properly, we obtain,
$\Rightarrow \left[ {15{\mkern 1mu} (t - 8)} \right] = \left[ {120 - 60} \right]$
$\Rightarrow t - 8 = \dfrac{{60}}{{15}}$
Upon rearranging the equation we get,
$\Rightarrow t = 8 + 4$
$\Rightarrow t = 12\;\sec$
It is given that this time is equal to $2\;x$ .
Therefore $2x = 12\;\sec$ .
$\Rightarrow x = 6$
Thus the total time taken by the motorcycle to travel the entire distance of $120\;m$ is $12\;\sec$ with the value of $x$ being $6$ .

Note:
The motorcycle moves with some acceleration in the first half of the motion. In the second half of the motion, the acceleration of the motorcycle is $0$ . Since the two graphs are two separate straight lines, we have analyzed them differently and have used the major relation $v = \dfrac{{ds}}{{dt}}$ to simplify and integrate to get the solution.