A car travels the first $\dfrac {2}{5}$th part of its total distance with a speed v₁, and the remaining $\dfrac {3}{5}$th of the distance with speed v₂. Its average speed is :

1. $\dfrac{1}{2}\sqrt{\text v$1 \text v2}

2. $\dfrac{\text v$1 +\text v2}{2}

3. $\dfrac{2\text v$1 \text v2}{\text v1 +\text v2}

4. $\dfrac{5\text v$1 \text v2}{3\text v1 +2\text v2}

$\dfrac{5\text v$1 \text v2}{3\text v1 +2\text v2}

Reason — Let, total distance covered be S of which S₁ is covered with v₁ in time t₁ and S₂ with v₂ in time t₂.

Then,

$\text S_1=\dfrac{2}{5}\text S$ …………… (1)

and

$\text S_2=\dfrac{3}{5}\text S$ …………… (2)

As

$\text {Speed (v)}=\dfrac{\text {Distance (S)}}{\text {Time (t)}}$

Then,

$\Rightarrow \text v$1=\dfrac{\text S1}{\text t1} \\[1 em] \Rightarrow \text t1=\dfrac{\text S1}{\text v1}

Putting value of S₁ from equation 1

$\Rightarrow \text t$1 = \dfrac{2}{5}\dfrac{\text S}{\text v1}

Similarly

$\Rightarrow \text v$2=\dfrac{\text S2}{\text t2} \\ [1 em] \Rightarrow\text t2=\dfrac{\text S2}{\text v2}

Putting value of S₂ from equation 2

$\Rightarrow\text t$2= \dfrac{3}{5}\dfrac{\text S}{\text v2} Now,

$\text {Average Speed (v)}=\dfrac{\text {Total distance travelled (S)}}{\text {Total time taken (t)}} \\[1 em] =\dfrac{\text S}{\text t$1 + \text t2} \\[1 em] =\dfrac{\text S}{\dfrac{2\text S}{5\text v1}+\dfrac{3\text S}{5\text v2}} \\[1 em] = \dfrac{\text S}{\dfrac{\text S}{5}\Big(\dfrac{2}{\text v1} + \dfrac{3}{\text v2}\Big)} \\[1em] = \dfrac{\text S}{\dfrac{\text S}{5}\Big(\dfrac{2\text v2 + 3 \text v1}{\text v1 \text v2} \Big)} \\[1em] =\dfrac{5\text v1 \text v2}{3\text v1 +2\text v2}