The Application of Differential Equations in Physics

Differential equations are commonly used in physics problems. In the following example we shall discuss a very simple application of the ordinary differential equation in physics.

Example:
A ball is thrown vertically upward with a velocity of 50m/sec. Ignoring air resistance, find

(i) The velocity of the ball at any time t
(ii) The distance traveled at any time t
(iii) The maximum height attained by the ball

Let v and h be the velocity and height of the ball at any time t. Since the time rate of velocity is acceleration, so \frac{{dv}}{{dt}} is the acceleration. Since the ball is thrown upwards, its acceleration is  - g. Thus, we have

\frac{{dv}}{{dt}} = - g\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Separating the variables, we have

dv = - gdt\,\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right)

(i) Since the initial velocity is 50m/sec, to get the velocity at any time t, we have to integrate the left side (ii) from 50 to v and its right side is integrated from 0 to t as follows:

\begin{gathered} \int_{50}^v {dv = - g\int_0^t {dt} } \\ \Rightarrow \left| v \right|_{50}^v = - g\left| t \right|_0^t \\ \Rightarrow v - 50 = - g\left( {t - 0} \right) \\ \Rightarrow v = 50 - gt\,\,\,\,{\text{ - - - }}\left( {{\text{iii}}} \right) \\ \end{gathered}

Since g = 9.8m/{s^2}, putting this value in (iii), we have

v = 50 - 9.8t\,\,\,\,{\text{ - - - }}\left( {{\text{iv}}} \right)

(ii) Since the velocity is the time rate of distance, then v = \frac{{dh}}{{dt}}. Putting this value in (iv), we have

\frac{{dh}}{{dt}} = 50 - 9.8t\,\,\,\,{\text{ - - - }}\left( {\text{v}} \right)


Separating the variables of (v), we have

dh = \left( {50 - 9.8t} \right)dt\,\,\,\,\,{\text{ - - - }}\left( {{\text{vi}}} \right)

In order to find the distance traveled at any time t, we integrate the left side of (vi) from 0 to h and its right side is integrated from 0 to t as follows:

\begin{gathered} \int_0^h {dh} = \int_0^t {\left( {50 - 9.8t} \right)dt} \\ \Rightarrow \left| h \right|_0^h = \left| {50t - 9.8\frac{{{t^2}}}{2}} \right|_0^t \\ \Rightarrow h - 0 = 50t - 9.8\frac{{{t^2}}}{2} - 0 \\ \Rightarrow h = 50t - 4.9{t^2}\,\,\,\,\,{\text{ - - - }}\left( {{\text{vii}}} \right) \\ \end{gathered}

(iii) Since the velocity is zero at maximum height, we put v = 0 in (iv)

\begin{gathered} 0 = 50t - 9.8{t^2} \Rightarrow 0 = 50 - 9.8t \\ \Rightarrow t = \frac{{50}}{{9.8}} = 5.1 \\ \end{gathered}

Thus, the maximum height is attained at time t = 5.1\,\sec .

Putting this value of t in equation (vii), we have

\begin{gathered} h = 50\left( {5.1} \right) - 4.9{\left( {5.1} \right)^2} \\ \Rightarrow h = 255 - 127.449 = 127.551 \\ \end{gathered}

Thus the maximum height attained is 127.551{\text{m}}.