Question

A car covers 4 successive 3 km stretches at speeds of 10kmph, 20 kmph, 30 kmph and 60 kmph

respectively. Its average speed is​

Answer 1

Case 1

Distance =4km

Speed = 10 kmph

Time = Distance/speed

Time taken =

$= \frac{4}{10} \\ \\ = \frac{2}{5} hr$

$\rule{200}{2}$

Case 2:-

Distance = 4km

Speed= 20 kmph

Time =Distance/speed

Time = 4/20 = 1/5 hr

$\rule{200}{2}$

Case 3 :-

Distance = 4km

Speed = 30kmph

Time = Distance /speed

Time taken =

$= \frac{4}{30} \\ \\ = \frac{2}{15}hr$

$\rule{200}{2}$

Case 4

Distance = 4km

Speed = 60kmph

Time = Distance /speed

Time = 4/60 = 1/15hr

$\rule{200}{2}$

Now,

Total distance travelled will be :-

$= 4 + 4 + 4 + 4 \\ \\ = 4 \times 4 \\ \\ = 16km$

$\rule{200}{2}$

Total time taken will be :-

$= \frac{2}{5} + \frac{1}{5} + \frac{2}{15} + \frac{1}{15} \\ \\ \\ = \frac{6 + 3 + 2 + 1}{15} \\ \\ \\ = \frac{12}{15} \\ \\ \\ = \frac{4}{5}hr$

$\rule{200}{2}$

Average speed = Total distance /Total time

Hence,

Average speed is equal to

$\frac{16}{ \frac{4}{5} } \\ \\ \\ = \frac{16 \times 5}{4} \\ \\ \\ = 4 \times 5 \\ \\ = 20kmph$

$\rule{200}{2}$

Answer 2

Distance=Speed×Time

Time=Distance÷Speed

Speed=Distance÷Time

1st strech 3 km at=10 km/h

Time=3÷10 hours

$= \frac{3}{10} \: hours$

2nd strech 3km at =20 km/h

Time=3÷20

$= \frac{3}{20} \: hours$

3rd strech 3 km at=30 km/h

Time=3÷30

$= \frac{3}{30} \\ \\ = \frac{1}{10} \: hours$

4th Strech 3 km at=60 km/h

Time=3÷60

$= \frac{3}{60} \\ \\ = \frac{1}{20} \: hours$

Total Distance=1st strech+2nd strech +3rd strech+4th strech

=3+3+3+3

=12 km

Total Distance=12 km

Total Time

$= \frac{3}{10} + \frac{3}{20} + \frac{1}{10} + \frac{1}{20} \\ \\ = \frac{2 \times 3 + 3 + 2 + 1}{20} \\ \\ = \frac{6 + 3 + 2 + 1}{20} \\ \\ = \frac{12}{20} \\ \\ = \frac{3}{5} \: hours$

$\boxed{\mathtt{\huge{Average\:Speed=\frac{Total\: Distance}{Total\:Time}}}}$

$\frac{12}{ \frac{3}{5} } \\ \\ = \frac{12}{3} \times 5 \\ \\ = 4 \times 5 \\ \\ = 20 \:$

$\underline{\mathtt{\huge{Average\:Speed=20\:km/h}}}$