Question

Free Fall When a rock falls from rest near the surface of the earth, the distance it covers during the first few seconds is given by the equation $$s=4.9 t^{2}$$ In this equation, \(s\) is the distance in meters and \(t\) is the elapsed time in seconds. Find \(d s / d t\) and \(d^{2} s / d t^{2}\)

Short answer

The first derivative \(ds/dt = 9.8t\) represents the velocity of the falling rock. The second derivative \(d^2s/dt^2 = 9.8\) represents the acceleration due to gravity.

Step-by-step solution

Find the first derivative

The first derivative of \(s\) with respect to \(t\) (denoted as \(ds/dt\)) will give the rate of change of \(s\) with respect to \(t\), which represents the velocity in this context. Using the power rule for differentiation, which states that the derivative of \(t^n\) is \(n*t^{(n-1)}\), the first derivative, \(ds/dt\), can be found. Apply the rule to the equation \(s = 4.9t^2\) to obtain: \(ds/dt = 2 * 4.9t = 9.8t\).

Find the second derivative

The second derivative of \(s\) with respect to \(t\) (denoted as \(d^2s/dt^2\)) will give the rate of change of the velocity with respect to time, which represents the acceleration in this context. For this, differentiate \(ds/dt\) to get \(d^2s/dt^2\). The derivative of \(9.8t\) is simply \(9.8\), so \(d^2s/dt^2 = 9.8\). This value represents the acceleration due to gravity.