Question

A car travels the first third of a distance with a speed of 10 kmph, the
second third at 20 kmph and the last third at 60 kmph. What is its me
speed over the entire distance?
(Ans: 0.56 m​

Answer 1

Given

A car travels the first third of a distance with a speed of 10 km/h, the  second third at 20 km/h and the last third at 60 km/h

To Find

Speed for entire distance ( Average speed )

Concept Used

Average speed is defined as total distance covered per total time

$\bf \pink{\bigstar\ \; A_{Speed}=\dfrac{T_{Distance}}{T_{time}}}$

Speed is defines distance traveled per unit time

$\bf \green{\bigstar\ \; Speed=\dfrac{Distance}{Time}}$

Solution

Let the total distance = x km

Case 1 : For 1st third of distance

Distance = ˣ/₃ km

Speed = 10 km/h

$\to \rm Speed=\dfrac{Distance}{Time}\\\\\to \rm t_1=\dfrac{\frac{x}{3}}{10}\\\\\to \bf t_1=\dfrac{x}{30}$

Case - 2 : For 2nd third of distance

Distance = ˣ/₃ km

Speed = 20 km/h

$\to \rm t_2=\dfrac{\frac{x}{3}}{20}\\\\\to \bf t_2=\dfrac{x}{60}$

Case - 3 : For 3rd third of distance

Distance = ˣ/₃ km

Speed = 60 km/h

$\to \rm t_3=\dfrac{\frac{x}{3}}{60}\\\\\to \bf t_3=\dfrac{x}{180}$

Now , apply formula for average speed ,

$\implies \bf A_{Speed}=\dfrac{T_{Distance}}{T_{time}}\\\\\implies \rm A_{Speed}=\dfrac{x}{\frac{x}{30}+\frac{x}{60}+\frac{x}{180}}\\\\\to \rm A_{Speed}=\dfrac{x}{\frac{3x}{60}+\frac{x}{180}}\\\\\to \rm A_{Speed}=\dfrac{x}{\frac{10x}{180}}\\\\\to \rm A_{Speed}=\dfrac{180x}{10x}\\\\\to \bf A_{Speed}=18\ km/h$

Answer 2

Answer:

V1 is equals to 10 km per hour V2 is equals to 20 km with 60 km mean speed is equals to 3 v1 V2 V3 by V1 V2 + V2 V3 + V3 V1 that is equals to 3 into 10 into 20 into 60 by 10 into 20 + 20 x 60 + 60 * 10 that is equals to 30 into 1200 by 200 + 1200 + 600 that is equals to that is equals to 36000 by 2000 is equals to 18 main speed is equals to 36000 by 2000 cancel is equals to 18 km is equals to 18 into 5 by 8 metre per second is equals to 5 m per second inverse