Find the Average speed of a man who travels different distances in different time as given below :
| Distance | Time |
| \[10\] km | \[1\] hr |
| \[20\] km | \[2\] hr |
| \[30\] km | \[3\] hr |
If a man travels different distances \[d_1, d_2, d_3,.....\] and so on in different time \[t_1,t_2,t_3,.....\] respectively then, Average speed
\[=\frac{\text{total}\ \text{distance}}{\text{total}\ \text{time}}\]
\[=\frac{d_1+d_2+d_3+.....}{t_1+t_2+t_3+.....}\]
\[\therefore\] Average speed of the man
\[=\frac{10+20+30}{1+2+3}\]
\[=\frac{60}{6}\]
\[=10\] km/h
Find the Average speed of a man who travels different distances with different speed as given below :
| Distance | Speed |
| \[10\] km | \[1\] km/h |
| \[20\] km | \[2\] km/h |
| \[30\] km | \[3\] km/h |
If a man travels different distances \[d_1, d_2, d_3,.....\] and so on with different speed \[s_1,s_2,s_3,.....\] respectively then, Average speed
\[\frac{\left(d_1+d_2+d_3+.....\right)}{\frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}+.....}\]
\[ \therefore \text{Requried}\ \text{Distance}\]
\[=\frac{\left(10+20+30\right)}{\frac{10}{1}+\frac{20}{2}+\frac{30}{3}}\]
\[=\frac{60}{10+10+10}\]
\[=\frac{60}{30}\]
\[=2\] hour
A bus travels from A to B with the speed \[30\] km/h and returns from B to A with the speed \[60\] km/h. What is the average speed of the bus?
If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h. then the average speed will be \[\frac{2xy}{x+y}\]
\[=\frac{\left(2\times 30\times 60\right)}{30+60}\]
\[=\frac{3600}{90}\]
\[=40\] km/h
A man covers \[\frac{1}{8}\] part of Journey at \[2\] km/h, \[\frac{1}{2}\] part at \[13\] km/h. Find his average speed for the whole journey?
If a man covers \[\frac{1}{x}\] part of Journey at u km/h, \[\frac{1}{y}\] part at v km/h. then his average speed for the whole journey will be
\[\frac{xu\times yv}{xu+yv}\]
\[\therefore\] Average speed
\[=\frac{\left(2\times 6\right)\times \left(4\times 12\right)}{\left(2\times 6\right)+\left(4\times 12\right)}\]
\[=\frac{24\times 26}{50}\]
\[=\frac{624}{50}\]
\[=12.48\] km/h
If anyone overtakes or follows another, then the time taken to catch
\[= \frac{\text{Distance between them}}{\text{Relative speed}}\]
According to the question,
The distance between them \[= 90\] meter
Relative speed \[= 2\] km/hr
\[=\left(2\times \frac{5}{18}\right)\] m/sec
=\[\frac{5}{9}\] m/sec
\[\therefore\] Required time
\[= \frac{\text{Distance between them}}{\text{Relative speed}}\]
\[\frac{90}{\frac{5}{9}}\] sec
\[=\frac{90\times 9}{5}\] sec
\[=18\times 9\] sec
\[=162\] sec
If anyone overtakes or follows another,
Total traveled distance to catch the slower one
\[= \frac{\text{(Product of speeds)} \times \text{time}}{\text{(Difference of speeds)}}\]
\[\frac{\left(60\times 40\right)\times \frac{2}{60}}{60-40}\]
\[=\frac{2400\times \frac{1}{30}}{20}\]
\[=\frac{80}{20}\]
\[=4\] km
The formula to calculate the distance traveled by the train is
Distance \[= (n – 1)x\]
where
\[\therefore\] The distance traveled between the train
\[= (n – 1)x\]
\[= (21-1)100\] meters
\[= 20 \times 100\] meters
\[= 2000\] meters
\[= \frac{2000}{1000}\] km
\[=2\] km
A train running at a speed of \[72\] km/hr crosses an electric pole in \[2\] minutes. What is the length of the train?
The formula to calculate the length of a train is
\[\text{Length of train} = \text{Speed} × \text{Time}\]
According to the question
The speed of train \[=72\times \frac{5}{18}=20\] m/s
Time to cross the pole \[=2\] minutes \[= 2 × 60 = 120\] seconds
\[\therefore \text{Length of train} = \text{Speed} × \text{Time}\]
\[=20 × 120 = 2400\] meters
\[=\frac{2400}{1000}=2.4\] km
Let a metre long train is going with the speed x m/s and b metre long train is also going with the speed y m/s in the same direction on parallel path,
Then total time taken by the faster train to cross the slower train
\[=\frac{a+b}{x-y}\] seconds
\[\therefore\] Required time
\[=\frac{1500+1000}{10-5}\] seconds
\[=\frac{2500}{5}\] seconds
\[=500\] seconds
Let a metre long train is traveling with the speed x m/s and b metre long train is travelling with the speed y m/s in the opposite direction on parallel path.
Then, time taken by the trains to cross each other
\[=\frac{a+b}{x+y}\] seconds
\[\therefore\] Required time
\[=\frac{500+250}{10+5}\] seconds
\[=\frac{750}{15}\] seconds
\[=30\] seconds
A train crosses a standing pole in \[20\] sec time and crosses a \[50\] metre long platform in \[60\] sec. What is the length of the train
If a train crosses a standing man/a pole in ‘\[t_1\]’ sec time and crosses P meter long platform in ‘\[t_2\] ’ sec. time,
Then the length of the train
\[=\frac{P \times t_1}{t_2 - t_1}\]
\[=\frac{50 \times 20}{60-20}\]
\[=\frac{50 \times 20}{40}\]
\[=\frac{1000}{40}\]
\[=250\] metre
A \[400\] metre long train is running with the speed \[10\] m/s. A man is running in the same direction and with the speed \[2\] m/s. What is the time taken by the train to cross the man?
Let a metre long train is running with the speed x m/s. A man is running in the same direction and with the speed y m/s,
Then the time taken by the train to cross the man
\[=\frac{a}{(x-y)}\] seconds
\[=\frac{400}{(10-2)}\] seconds
\[=\frac{400}{8}\] seconds
\[=50\] seconds
A train crosses two men in \[20\] seconds and \[10\] seconds running in the same direction with the speed \[2\] m/s and \[1\] m/s. Find the speed of train.
If a train crosses two men in \[t_1\] seconds and \[t_2\] seconds running in the same direction with the speed \[s_1\] and \[s_2\] .
Then the speed of the train is.
\[\frac{t_1s_1 - t_2s_2}{t_1 - t_2}\]
\[\therefore\] Required speed
\[=\frac{(20 \times 2) - (10 \times 1)}{20-10}\]
\[=\frac{(40) - (10)}{10}\]
\[=\frac{30}{10}\]
\[=3\] m/s
Two trains of (same lengths) are coming from same direction and cross a man in \[20\] and \[10\] seconds. Find the time taken by both the trains to cross each other.
If two trains of (same lengths) are coming from the same direction and cross a man in \[t_1\] and \[t_2\] seconds.
Then the time taken by both the trains to cross each other is
\[\frac{2 ́ \times \text{Product of time}}{\text{Difference of time}}\]
\[=\frac{2 ́ \times t_1 \times t_2}{t_1 - t_2}\]
\[=\frac{2 ́ \times (20 \times 10)}{20-10}\]
\[=\frac{2 ́ \times (200)}{10}\]
\[\frac{400}{10}\]
\[=40\] seconds
Two trains of the same length are coming from opposite directions and cross a man in \[5\] seconds and \[10\] seconds. Find the time taken by both trains to cross each other.
If two trains of the same length are coming from opposite directions and cross a man in \[t_1\] seconds and \[t_2\] seconds.
Then the time taken by both trains to cross each other
\[\frac{2 \times \text{Product of time}}{\text{Sum of time}}\]
\[=\frac{2 \times 4 \times 6}{4 + 6}\]
\[=\frac{48}{10}\]
\[=4.8\] seconds
A train of length \[300\] m crosses a platform of length \[50\] m with the speed \[7\] m/s. Find the time taken by the train to cross the platform.
If a train of length x m crosses a platform of length y m with the speed u m/s in t seconds, then
\[t=\frac{x+y}{u}\]
\[t=\frac{300+50}{7}\]
\[t=\frac{350}{7}\]
\[=50\] seconds
From stations A and B, two trains start traveling towards each other at speeds \[15\] m/s and \[10\] m/s, respectively. When they meet each other, it was found that one train covers distance \[300\] metre more than that of another train. Find the distance between stations A and B.
From stations A and B, two trains start traveling towards each other at speeds a and b, respectively. When they meet each other, it was found that one train covers distance d more than that of another train. The distance between stations A and B is given as.
\[\frac{a+b}{a-b} \times d\]
\[\therefore\] Required distance
\[=\frac{15+10}{15-10} \times 300\]
\[=\frac{25}{5} \times 300\]
\[=5 \times 300\]
\[=1500\] metre