Question

The velocity graph of an accelerating car is shown.

(a) Estimate the average velocity of the car during the first \(12\) seconds.

(b) At what time was the instantaneous velocity equal to the average velocity?

Short answer

(a) Average velocity is\(45\).

(b) It will be equal at \(4.5\)seconds.

Step-by-step solution

Graph of the function.

Average velocity.

(a) Now, find the average velocity of the car during the first\(12\)seconds.

The average value of a function is represented by this graph is,

\({{\rm{f}}_{{\rm{axe}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{b - a}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{f(}}{{\rm{x}}_{\rm{i}}}{\rm{)\Delta x}}} \)

Here \(\Delta x = \frac{{b - a}}{n}\)and\(n\)is a number of terms.

From the graph, \(a = 0,b = 12\)

Let us take \(n = 6\) and

\(\begin{aligned}{c}{\rm{\Delta x &= }}\frac{{{\rm{b - a}}}}{{\rm{n}}}\\{\rm{ &= }}\frac{{{\rm{12 - 0}}}}{{\rm{6}}}\\{\rm{ = 2}}\end{aligned}\)

Average value.

Now, from the graph the average value is

\(\begin{aligned}{c}{{\rm{f}}_{{\rm{axe}}}}{\rm{ &= }}\frac{{\rm{1}}}{{{\rm{b - a}}}}\sum\limits_{{\rm{i &= 1}}}^{\rm{n}} {{\rm{f'(}}{{\rm{x}}_{\rm{i}}}{\rm{)\Delta x}}} \\{\rm{ &= }}\frac{{\rm{1}}}{{{\rm{12 - 0}}}}\left( {{\rm{\Delta x}}\left( {{\rm{f(1) + f}}\left( {\rm{3}} \right){\rm{ + f}}\left( {\rm{5}} \right){\rm{ + f}}\left( {\rm{7}} \right){\rm{ + f}}\left( {\rm{9}} \right){\rm{ + f}}\left( {{\rm{11}}} \right)} \right)} \right)\\{\rm{ &= }}\frac{{\rm{1}}}{{{\rm{12}}}}{\rm{.2}}\left( {{\rm{10 + 30 + 45 + 55 + 65 + 65}}} \right)\\{\rm{ &= }}\frac{{\rm{1}}}{{\rm{6}}}{\rm{.270}}\\{\rm{ &= 45}}\end{aligned}\)

Therefore, the average velocity is\(45\).

Instantaneous velocity.

(b) From the graph,

The velocity of a car was\(45km/h\)at \(4.5\)seconds.

Therefore, instantaneous velocity is equal to average velocity at \(4.5\)seconds.

\(t = 4.5\).