Time and Distance

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Time and Distance rules –

(1) Speed= Distance / Time

(ii) Distance = Speed × Time

(iii) Time= Distance / Speed

(iv) If the speed of a body is changed in the ratio a: b, then the ratio of

the time occupied variations in the ratio b: a.

(v) x km/hr = (X × 5/18 ) m/sec.

(vi) x metres/sec = (X × 18/5 ) km/hr.

 

Ex 1: Express a speed of 18 km/hr in metres per second.

Soln: 18 km/hr = ( 18  × 5/18 ) m/sec = 5 meter/sec

 

Ex 2: Express 10 m/sec in km/hr.

Soln: 10 m/sec  = ( 10   × 18/5 ) km/hr = 36 km/hr

 

Theorem: If a convinced distance is sheltered at x km/hr and the same distance is covered at y km/hr then the average speed during the entire journey is 2xy / x +y km/hr.

 

Proof: Let the distance be A km.

Time taken to travel the distance at a speed of x km/hr = A /x hrs.

Time occupied to travel the distance at a speed of y km/hr = A /y hrs.

Thus, we see that the total distance of 2A km is travelled in (A / x + A / y ) hrs

average speed = 2A / A / x = A / Y = 2Axy / A(x+y) = 2xy / x+y km/hr.

 

Ex. 3: A man covers a certain distance by car driving at 70 km/hr and he returns back to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.

Soln: Average speed = 2×70×55 / 70+55 km/hr = 61.6 km/hr.

 

Ex. 4: A man covers a convinced distance between his house and office on scooter. He is delayed by 10 minutes, despite having an average speed of 30 km/hr..Though, with a speed of 40 km/hr, he ranges his office 5 min earlier. Find the distance between his house and office.

Soln: Let the distance be x km.

Time taken to cover x km at 30 km/hr = x / 30hrs.

Time taken to cover x km at 40 km/hr = x / 40 hrs.

Alteration between the time taken = 15 min = 1 / 4 hr.

X/30 – x/40 = ¼ or 4x – 3x = 30 or x = 30

Hence, the required distance is 30 km.

Direct formula: Required distance = Product of two speeds / Difference of two speeds × Difference between arrival times.

 

Ex. 5: A man walking with a speed of 5 km/hr reaches his goal 5 minutes late. If he walks at a speed of 6 km/hr, he ranges on time. Find the distance of his target from his house.

Soln: This is similar to Ex. 4. Now the alteration in time is 5 minutes only.

Thus, required distance = 5 × 6 / 6 – 5 × 5 / 60 = 5/2 km = 2.5 km

 

Ex. 6: A boy walking at a speed of 10 km/hr reaches his school 15 minutes late. Afterwards, he augments his velocity by 2 km/hr, yet he remains delayed by 5 minutes. Determine the distance between his residence and his educational institution.

Soln: Here, the difference in time = 15-5= 10 minutes.

= 1 / 6 hours

His speed during next journey =  10+2  = 12 km/hr.

required distance = 12 × 10 / 12 – 10 × 1/6 = 10 km

 

Ex. 7: A boy goes to school at a speed of 3 km/hr and returns to the village at a speed of 2 km/hr. If he takings 5 hrs in all, what is the distance among the village and the school?

Soln: Let the required distance be x km.

Then time taken during the first journey = x/3 hr

and period occupied during the second journey = x/2 hr

x/3 + x/2 = 5 = 2x + 3x / 6 = 5 = 5x = 30

x = 6

required distance = 6 km

Direct formula: Required distance =  Product of the two speeds / Addition of the two speeds

=  5 x 3×2 / 3+2  = 6 km

 

Ex. 8: A motor car prepares a voyage in 10 hrs, the first half at 21 km/hr and

the second half at 24 km/hr. Find the distance.

Soln: This question is similar to Ex. 7, but we can’t use the direct formula

(used in Ex. 7) in this case. If we use the above formula, we get half of the distance. (But why?) See the detailed method first.

Let the distance be x km.

Then x/2km is travelled at a speed of 21 km /hr and x/2  km at a speed of 24km/hr.

Then time taken to travel the whole journey

= x/2 × 21 + x / 2 × 24 = 10 hrs

So x = 2×10×21×24 / (21+24) = 224 km

 

Ex. 9: The distance between two stations, Delhi and Amritsar, is 450 km. A train starts at 4 p.mA train departs from Delhi and travels towards Amritsar with an average velocity of 60 km/hr. Meanwhile, another train commences its journey from Amritsar at 3.20 p.m., heading towards Delhi with an average speed of 80 km/hr. How far-off from Delhi will the two trains encounter and at what time ?

Soln: Suppose the trains meet at a distance of x km from Delhi. Occupancy the trains from Delhi and Amritsar be A and B correspondingly. Then, [Time occupied by B to concealment (450 – x) km]

-[Time taken by A to cover x km] = 40/60 (see note)

450x – x / 80 – x/60 = 40/60

3  (450-x)-4x= 160 => 7x= 1190 ⇒ x=170

Therefore, the trains encounter at a distance of 170 km from Delhi.

Time taken by A to cover 170 km = (170/60)hrs=2hrs 50 min

So, the trains encounter at 6.50 p.m.

Note: RHS= 4:00 p.m. – 3:20 p.m. = 40 minutes = 40hr / 60 hr

LHS comes from the fact that the train from Amritsar took 40 minutes more to travel up to the meeting point because it had started its journey at 3.20 p.m. whereas the train from Delhi had started its journey at 4 p.m. and the meeting time is the same for both the trains.

 

Ex. 10: Walking 4th of his usual speed, a person is 10 min late to his

office. Find his typical time to concealment the distance.

Soln: Let the usual time be x min.

Time taken at ¾ th  of the usual speed = 4x/3 min (from (iv) under formulae  section)

4/3x – x = 10 = x/3 = 10 = x = 30 min

 

Ex. 11: Running 4/3th of his usual speed, a person improves his timing by 10 minutes. Determine his regular timeframe for completing the distance.

Soln: This is similar to Ex 10, but not exactly the same. In this case, the speed is increased and hence the time is reduced. Whereas, it was just opposite in Ex. 10.

You should try to solve this question by detailed method.

Direct formula for such question is slightly changed and is given as:

Usual time =  Improved time / 1 – 1÷ 4/3 = 10/1-3/4 = 40 minutes

Note: Mark the change in the above two direct formulae.

 

 

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