Question

A car travels with speed V, 2V, 3V.... nV for successive time intervals which are in the ratio 1:2:3....n find the average speed.

Answer 1

Given:

Car travels with speeds v , 2v , 3v .... nv for successive time intervals in the ratio of 1:2:3....

To find:

Average speed of car

Concept:

Average Speed is defined as the ratio of total distance Travelled to the total Time taken to travel that distance .

Its a scalar quantity , having only magnitude and no direction.

Calculation:

For time , let t be the constant of proportionality.

$v \: avg. = \dfrac{total \: distance}{total \: time}$

$= > v \: avg. = \dfrac{(v \times t) + (2v \times 2t) + .... + (nv \times nt)}{t + 2t + 3t + ... + nt}$

$= > v \: avg. = \dfrac{vt + 4vt + 9vt + .... + {n}^{2} vt}{t + 2t + 3t + .... + nt}$

$= > v \: avg. = \dfrac{v \cancel t(1 + 4 + 9 + .... + {n}^{2}) }{ \cancel t (1+ 2 + 3+ .... + n)}$

In the numerator , we have sum of squares , and in the denominator we have sum of natural numbers :

$= > v \: avg. = \dfrac{v \cancel t( {1}^{2} + {2}^{2} + {3}^{2} + .... + {n}^{2}) }{ \cancel t (1+ 2 + 3+ .... + n)}$

$= > v \: avg. = \dfrac{ \frac{n(n + 1)(2n + 1)}{6} }{ \frac{n(n + 1)}{2} }$

$= > v \: avg. = \dfrac{ \frac{ \cancel{n(n + 1)}(2n + 1)}{6} }{ \frac{ \cancel{n(n + 1)}}{2} }$

$= > v \: avg. = \dfrac{2n + 1}{3}$

So final answer is :

$\boxed{ \blue{ \bold{ \huge{v \: avg. = \dfrac{2n + 1}{3} }}}}$