Question

For the motion of a particle, velocity 'v' depends on displacement x as \[v = \dfrac{{20}}{{3x - 2}}\].If at t=0,x=0 then at what time t, the x=20?

Answer 1

Hint: We introduce calculus to differentiate the displacement together with the velocity and to differentiate the velocity to get the acceleration. Differentiation gives the rate of change of something. Also in other ways by integrating, we can integrate acceleration to get velocity and also integrate velocity to get the displacement. As we know that velocity is the rate of change of displacement known as velocity.
\[v = \dfrac{{dx}}{{dt}}\]
Now by arrange this equation as
\[dx = vdt\]
So the terms dt and dx need to be integrating as
\[\int {dx = \int {vdt} } \]
Here the limits on the dx integral are related to the displacement ‘x’ and the limits on the dt integral are related to the time ‘t’. After placing the limits on the integration we can find an expression for the displacement of the particle at any time.

Formula used Velocity is rate of change of displacement
\[v = \dfrac{{dx}}{{dt}}\]
Where v is velocity, x is displacement and t is time taken.
Some integral identity
\[\int {xdx = \dfrac{{{x^2}}}{2}} \]
\[\Rightarrow \int {dx = x} \]

Complete step by step solution:
For a motion of a particle, velocity and displacement relation is
\[v = \dfrac{{20}}{{3x - 2}}\]
At x=0,t=0 which means when displacement is zero then time also becomes zero.
Now by the given equation, we get;
\[v = \dfrac{{20}}{{3x - 2}}\]
Also,
\[v = \dfrac{{dx}}{{dt}} = \dfrac{{20}}{{3x - 2}} \\ \]
\[(3x - 2)dx = 20dt\]
Integrating on both sides
\[\int {(3x - 2)dx} = \int {20dt} \]

Applying the limits from initial limit to final limit. Initially the displacement is given as x=0 at the starting point and final displacement as x=20. Similarly initially the time is given as 0. Now we find time t.
\[\int\limits_0^{20} {(3x - 2)dx} = \int\limits_0^t {20dt} \]
On integrating this, we get
\[\left[ {\dfrac{{3{x^2}}}{2} - 2x} \right]_0^{20} = 20t\]
Applying limits
\[\left[ {\dfrac{{3 \times 20 \times 20}}{2} - 2 \times 20} \right] = 20t \\ \]
\[\Rightarrow 560 = 20t \\ \]
\[\therefore t = 28\]
Therefore when displacement x=20 then the time taken by the moving particle is 20 seconds.

Hence when “x=20” the value of “t=28”.

Note: The velocity is obtained by instantaneously changing displacement with respect to time. In other words velocity is the derivative of displacement with respect to time.
\[v = \dfrac{{dx}}{{dt}}\]
After arranging it
\[\int {dx = \int {vdt} } \]
\[\Rightarrow x = \int {vdt} \]
We can use this formula in terms of many problems.