Question

Graph of position (x) vs inverse of velocity ($\frac{1}{v}$) for a particle moving on a straight line is as shown. Find the time (in sec) taken by the particle to move from x = 3 m to x = 15 m. x(m)

$\frac{1}{v}$(sm$^{-1}$)

Answer 1

60

Answer 2

The problem provides a graph of position (x) versus the inverse of velocity ($\frac{1}{v}$) for a particle moving on a straight line. We need to find the time taken by the particle to move from $x = 3$ m to $x = 15$ m.

1. Find the equation of the line: The graph is a straight line passing through two points: $( \frac{1}{v}_1, x_1 ) = (2, 3)$ and $( \frac{1}{v}_2, x_2 ) = (8, 15)$. Let $Y = \frac{1}{v}$ and $X = x$. The general equation of a straight line is $X = mY + c$. Calculate the slope ($m$): $m = \frac{x_2 - x_1}{\frac{1}{v}_2 - \frac{1}{v}_1} = \frac{15 - 3}{8 - 2} = \frac{12}{6} = 2$.

   Now, use one of the points (e.g., (2, 3)) to find the y-intercept ($c$): $3 = 2(2) + c$ $3 = 4 + c$ $c = -1$.

   So, the equation relating $x$ and $\frac{1}{v}$ is: $x = 2 \left(\frac{1}{v}\right) - 1$

2. Express $\frac{1}{v}$ in terms of $x$: From the equation derived above: $x + 1 = \frac{2}{v}$ $\frac{1}{v} = \frac{x+1}{2}$

3. Relate $\frac{1}{v}$ to time and position: We know that velocity $v = \frac{dx}{dt}$. Therefore, $\frac{1}{v} = \frac{dt}{dx}$.

4. Set up the integral for time: Substitute the expression for $\frac{1}{v}$ into the differential equation: $\frac{dt}{dx} = \frac{x+1}{2}$ $dt = \frac{x+1}{2} dx$

   To find the total time ($T$) taken to move from $x = 3$ m to $x = 15$ m, integrate $dt$ from $x=3$ to $x=15$: $T = \int_{3}^{15} \frac{x+1}{2} dx$

5. Evaluate the integral: $T = \frac{1}{2} \int_{3}^{15} (x+1) dx$ $T = \frac{1}{2} \left[ \frac{x^2}{2} + x \right]_{3}^{15}$ Now, substitute the limits of integration: $T = \frac{1}{2} \left[ \left( \frac{15^2}{2} + 15 \right) - \left( \frac{3^2}{2} + 3 \right) \right]$ $T = \frac{1}{2} \left[ \left( \frac{225}{2} + 15 \right) - \left( \frac{9}{2} + 3 \right) \right]$ To combine terms, find common denominators: $T = \frac{1}{2} \left[ \left( \frac{225}{2} + \frac{30}{2} \right) - \left( \frac{9}{2} + \frac{6}{2} \right) \right]$ $T = \frac{1}{2} \left[ \left( \frac{255}{2} \right) - \left( \frac{15}{2} \right) \right]$ $T = \frac{1}{2} \left[ \frac{255 - 15}{2} \right]$ $T = \frac{1}{2} \left[ \frac{240}{2} \right]$ $T = \frac{1}{2} (120)$ $T = 60$ seconds

The time taken by the particle to move from $x = 3$ m to $x = 15$ m is 60 seconds.

Explanation of the solution:

1. Determine the linear relationship between position (x) and inverse velocity (1/v) from the given graph. The equation is $x = 2(1/v) - 1$.
2. Rearrange the equation to express $1/v$ in terms of $x$: $1/v = (x+1)/2$.
3. Use the kinematic relation $dt = dx/v$, which implies $dt/dx = 1/v$.
4. Substitute the expression for $1/v$ into the differential equation: $dt = (x+1)/2 dx$.
5. Integrate this expression from the initial position $x=3$ m to the final position $x=15$ m to find the total time. The definite integral evaluates to 60 seconds.