# Time and Distance

A man in a train notices that he can count 21 telephone posts in one minute.If they are known to be 50 meters apart , then at what speed is the train travelling ?

Number of gaps between 21 telephone posts = 20.

Distance travelled in 1 minute = (50 × 20) m= 1000 m = 1 km.

∴ Speed = 60 km/hr.

An express train travelled at an average speed of 100 km/hr,stopping for 3 minutes after every 75 km.How long did it take to reach its destination 600 km from the starting point ?

Time taken to cover 600 km = $$\left ( \frac{600}{100} \right )$$ hrs = 6 hrs.

Number of stoppages = $$\left ( \frac{600}{75} - 1 \right )$$ = 7

Total  time  of stoppages  = (3 × 7) min= 21 min.

Hence,total time taken = 6 hrs 21 min.

A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time,the ratio of the speed of the jogger to that of the cyclist is :

Let the distance covered by, the cyclist be x and the time taken be y. Then,

Required ratio = $$\frac{\frac{1}{2}x}{2y} : \frac{x}{y} = \frac{1}{4} : 1$$ = 1 : 4

The speed of a car increases by 2 km after every one hour. If the distance travelled in the first one hour was 35 km, what was the total distance travelleld in 12 hours ?

Total distance travelled in 12 hours = (35 + 37 +39 + …… upto 12 terms ).

This is an A.P. with first term, a = 35, number of terms, n= 12, common difference, d = 2.

∴ Required distance = $$\frac{12}{2}\left [ 2 \times 35 + \left ( 12 - 1 \right ) \times 2 \right ]$$

= 6(70 + 22)

= 552 km.

A train covers a distance of 10 km in minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :

$$Speed = \left ( 10 \times \frac{60}{12} \right )$$ km/hr = 50 km/hr.

New speed = (50 - 5) km/hr = 45 km/hr.

∴  Time taken = $$\frac{10}{45}$$ hr

= $$\left ( \frac{2}{9} \times 60 \right )$$ min

= $$\frac{40}{3}$$ min

= 13 min 20 sec.

A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes.The length of the bridge is :

$$Speed =\left ( 5 \times \frac{5}{18} \right )$$  m/sec = $$\frac{25}{18}$$ m/sec.

Distance covered  in 15 minutes = $$\left ( \frac{25}{18} \times 15 \times 60 \right )$$ m = 1250 m.

How long will a boy take to run round a square field of side 35 metres, if he runs at the rate of 9 km/hr ?

Speed = 9 km/hr = $$\left ( 9 \times \frac{5}{18} \right ) m/sec = \frac{5}{2}$$ m/sec

Distance = (35 × 4) m = 140 m

∴ Time taken = $$\left ( 140 \times \frac{2}{5} \right )$$ sec = 56 sec

A car is running at a speed of 108 kmph. What distance will it cover in 15 seconds ?

Speed = 108 kmph = $$\left ( 108 \times \frac{5}{18} \right )$$ m/sec = 30 m/sec.

∴ Distance  covered in 15 sec = (30 × 15) m = 450 m.

A truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 km in 45 minutes. The ratio of their speeds is :

Ratio of speeds = $$\left ( \frac{550}{60} \times \frac{18}{5} \right ) : \left ( \frac{33}{45} \times 60 \right )$$

= 33 : 44

= 3 : 4.

The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is :

Then, $$8x = \frac{400}{4} = 100$$ ⇒ $$x = \left ( \frac{100}{8} \right ) = 12.5$$.