Question

A particle moves along a straight line with displacement \(s\left( t \right)\), velocity \(v\left( t \right)\), and acceleration \(a\left( t \right)\). Show that

\(a\left( t \right) = v\left( t \right)\frac{{dv}}{{ds}}\)

Explain the difference between the meanings of the derivatives \(\frac{{dv}}{{dt}}\) and \(\frac{{dv}}{{ds}}\).

Short answer

Both derivatives show the change in velocity with respect to time and displacement respectively.

Step-by-step solution

The acceleration in terms of velocity

The accelerationis equal to the rate of change of velocity. So, \(a\left( t \right) = \frac{d}{{dt}}\left( v \right)\). Solve the above equation as follows:

\(\begin{array}{c}a\left( t \right) = \frac{{dv}}{{dt}}\\ = \frac{{dv}}{{ds}} \cdot \frac{{ds}}{{dt}}\\ = \frac{{dv}}{{ds}} \cdot v\left( t \right)\\ = v\left( t \right) \cdot \frac{{dv}}{{ds}}\end{array}\)

Hence, it is proved.

The derivatives

Here, the derivative \(\frac{{dv}}{{dt}}\) represents the rate of change of velocity with respect to time and equivalent toacceleration. On the other hand, \(\frac{{dv}}{{ds}}\) represent the rate of change of the velocity with respect to the displacement.