Question

A particle covers 3/4th of total distance with speed v1 and next 1/4 with v2. Find the average speed of the particle?

Answer 1

Let the total distance covered is 'd' . Time taken for travelling 3d/4 with v_1 velocity
t_1 = 3d/4(v_1)
Similarly, t_2 = d/4 (v_2)
Average speed = distance travelled/time taken
=d/t_1+t_2
=d/(3d/4 (v_1)+d/4 (v_2))
=d(3v_2+v_1)/4v_1v_2

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Answer 2

Answer:

The total average speed of the particle is $\frac{4v_1v_2}{3v_2+v_1}$.

Explanation:

The average speed of the particle is given as,

$v_a_v_g=\frac{S}{T}$      (1)

Where,

vavg=average speed

S=total distance covered by the particle

T=total time taken by the particle

From the question we have,

x₁=$\frac{3S}{4}$

x₂=$\frac{S}{4}$

And the total time is given as,

$T=\frac{x_1+x_2}{v_1+v_2}$       (2)

By substituting the required values in equation (2) we get;

$T=\frac{\frac{3S}{4} +\frac{S}{4} }{v_1+v_2}=\frac{S}{4}( \frac{3}{v_1} +\frac{1}{v_2} )$

$T=\frac{S}{4} (\frac{3v_2+v_1}{v_1v_2} )$     (3)

By using equation (3) in equation (1) we get;

$v_a_v_g=\frac{S}{\frac{S}{4} (\frac{3v_2+v_1}{v_1v_2} )}=\frac{4v_1v_2}{3v_2+v_1}$

Hence, the total average speed of the particle is $\frac{4v_1v_2}{3v_2+v_1}$.