Question

1. an astronaut drops a rock off the edge of a cliff on the moon. the distance, ( d(t) ), in meters, the rock travels after ( t ) seconds can be modeled by the function ( d(t)=0.8 t^{2} ) what is the average rate of change in speed, in meters per second, of the rock between 5 and 10 seconds after it was dropped? interpret the rate in context.

Explanation:

Step1: Find the derivative of \(d(t)\)

The derivative of \(d(t)=0.8t^{2}\) using the power rule \((x^{n})^\prime = nx^{n - 1}\) is \(d^\prime(t)=1.6t\). This gives the speed function.

Step2: Calculate the average rate of change of \(d^\prime(t)\)

The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(a = 5\), \(b = 10\), and \(f(t)=1.6t\).

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Answer:

The average rate of change in speed is \(1.6\) meters per second. This means that, on average, the speed of the rock is increasing by \(1.6\) meters per second for each second in the time interval from \(5\) to \(10\) seconds after it was dropped.