Question

2. A car is traveling a road that includes two sides of a rectangle with the
sides a and b with a constant speed. What is the speed if the
magnitude of the average velocity of the car is v?

Answer 1

The speed of the car in terms of the average velocity is $\frac{v(a+ b) }{\sqrt{a^2 + b^2} }$

The given parameters;

• length of each side of the triangle = a and b
• average velocity = v

The average velocity of the car is defined as the rate of change of displacement.

$average \ velocity = \frac{displacement}{time}$

The displacement of the car along the rectangle is obtained by applying Pythagoras theorem to calculate the diagonal of the rectangle ;

c² = a² + b²

$c = \sqrt{a^2 + b^2}$

$v= \frac{\sqrt{a^2 + b^2} }{t} \\\\t = \frac{\sqrt{a^2 + b^2} }{v}$

The speed of the car is defined as the rate of change of distance.

$speed = \frac{distance}{time}$

The distance around the 2 sides of the rectangle is calculated as;

Distance =  a + b

$speed = \frac{a+b}{t}$

$speed = \frac{v(a+ b)}{\sqrt{a^2 + b^2} }$

Thus, the speed of the car is $\frac{v(a+ b) }{\sqrt{a^2 + b^2} }$

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