Question

If a body covers first half of its journey with uniform speed v1 and the second half of its journey with uniform speed v2 then the average speed is

Answer 1

Avg. Speed= (total distance)÷(total time)

t1=S/2 ÷ v1

t2=S/2 ÷v2

.°. avg. V = (S1+S2)÷(t1+t2)

=[(t1•v1)+(t2•v2)] ÷ (t1+t2)

=[(S/2÷v1 × v1) + (S/2÷v2 × v2)]

÷ [(s/2÷ v1)+(S/2÷ V2)]

= (2•v1•v2) ÷ (v1+V2)

Answer 2

Given that,

A body covers first half of its journey with uniform speed v₁ and the second half of its journey with uniform speed v₂.

To find,

The average speed of the body.

Solution,

The total distance covered divided by the total time taken is equal to the average speed of the body. Let the total distance is d i.e. d = d/2+d/2

Let t₁ and t₂ are time for first half and the second half of the journey. So,

$t_1=\dfrac{d}{2v_1}$, d and v₁ are distance and speed for the first half

And

$t_2=\dfrac{d}{2v_2}$, d and v₂ are distance and speed for the second half

Let V is the average speed. So,

$V=\dfrac{d_1+d_2}{t_1+t_2}$

Putting the values of t₁ and t₂ in above formula. So,

$V=\dfrac{d}{\dfrac{d}{2v_1}+\dfrac{d}{2v_2}}$

So,

$V=\dfrac{d}{\dfrac{d}{2}(\dfrac{1}{v_1}+\dfrac{1}{v_2})}\\\\V=\dfrac{2}{(\dfrac{1}{v_1}+\dfrac{1}{v_2})}\\\\V=\dfrac{2v_1v_2}{v_1+v_2}$

So, the average speed of the body is $\dfrac{2v_1v_2}{v_1+v_2}$.