Question

Suppose that an object is moving along the ground and that the distance (in centimeters) it has travelled after t seconds is given by the formula f(t)=t2 cm. Calculate the average speed during the 1 millionth of a second after t=3.

Answer 1

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\ -------------------------------\\\\ f(t)= t^2 \qquad \begin{cases} t_1=3\\ t_2=3+\frac{1}{1,000,000}\\ \qquad 3+0.000001\\ \qquad 3.000001 \end{cases}\implies \cfrac{f(3.000001)-f(3)}{3.000001-3} \\\\\\ \cfrac{9.000006000001~~-~~9}{0.000001}\implies \cfrac{0.000006000001}{0.000001}\implies 6.000001$

Answer 2

The average speed is the average rate of change of the function.

The average speed during the 1 millionth of a second after t=3 is 6 cm/s

The function is given as:

$\mathbf{f(t)=t^2}$

The average rate of a function is:

$\mathbf{f'(t)=\frac{f(b) - f(a)}{b - a}}$

1 millionth of a second after t=3 means that:

$\mathbf{a = 3}$

$\mathbf{b = 3.000001}$

So, we have:

$\mathbf{f'(t)=\frac{f(3.000001) - f(3)}{3.000001 - 3}}$

$\mathbf{f'(t)=\frac{f(3.000001) - f(3)}{0.000001}}$

Expand the numerator

$\mathbf{f'(t)=\frac{3.000001^2 - 3^2}{0.000001}}$

$\mathbf{f'(t)=\frac{0.000006}{0.000001}}$

$\mathbf{f'(t)=6 }$

Hence, the average speed during the 1 millionth of a second after t=3 is 6 cm/s

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