Question

Constant Acceleration Suppose the velocity of a falling body is \(v=k \sqrt{s} \mathrm{m} / \mathrm{sec}(k\) a constant) at the instant the body has fallen \(s\) meters from its starting point. Show that the body's acceleration is constant.

Short answer

The body's acceleration is constant, and equal to \(k/2\).

Step-by-step solution

Interpret the given equation

The velocity \(v\) at any instance during the fall depends on the distance fallen \(s\) and is given as \(v = k \sqrt{s}\), where \(k\) is a constant.

Differentiate the equation for velocity with respect to time

Since velocity \(v\) is the first derivative of distance \(s\) with respect to time \(t\), we have: \(v = ds/dt\). But from the original equation given, \(v = k \sqrt{s}\). Equating these two we get: \(ds/dt = k \sqrt{s}\). In other words, \(s = (ds/dt)^2 / k^2 = (v/k)^2\). Differentiate this equation w.r.t. time to obtain acceleration: \(a = d^2s/dt^2 = 2v/k * dv/dt = 2v/k * a\). Here, \(a = dv/dt\) as \(a\) is acceleration.

Solve for acceleration

Solving the last equation for \(a\) we get \(a = k/2\). This shows acceleration \(a\) is a constant, as it doesn't depend on \(v\), \(t\) or \(s\).