Question

A car travels 20km in first hour, 40km in second hour and 30km in third hour. calculate the average speed of the train?​

Answer 1

Method 1

$\bf{ \boxed{ \red{ \mathbb{average \: speed = \frac{total \: distance}{total \: time} }}}}$

according to the question,

Total distance in meters=(20+40+30)×1000m

=90,000m

total time in seconds=(1+1+1)×60×60sec=3×3600sec

=10,800sec

so,

$average \: speed = \frac{90000}{10800} = 8.333...meter \: per \: sec \\ \bf{ \boxed{ \green{ \mapsto{ \mathbb{ans : 8.34 \: meter \: per \: sec}}}}}$

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Method 2

Alternatively you can also find average speed mathematically,

$s_{1} = \frac{20000}{60 \times 60} = \frac{20000}{3600} m \: per \: sec \\ s_{2} = \frac{40000}{60 \times 60} = \frac{40000}{3600} m \:per \: sec \\ s_{3} = \frac{30000}{60 \times 60} = \frac{30000}{3600} m \: per \: sec$

Now,

$average \: speed = \frac{s_{1} +s_{2} + s_{3}}{3} \\ = \frac{20000 + 40000 + 30000}{3600 \times 3} = \frac{90000}{10800} \\ \bf{ \boxed{ \pink{ \mapsto{ \mathbb{ans : 8.34 \: meter \: per \: sec}}}}}$

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Hope it helps you....

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Answer 2

Distance in first hour =20km

Distance in second hour =40km

Distance in third hour =30km

•°• Average distance =

$\frac{20 + 20 + 40}{2} \\ \frac{90}{2} \\ = 45$