Time and Speed - Train and Boat

Relative speed = ( 45 + 30 )
= 75 km/hr
= (75 × 5)/18 m/s
= 125/6 m/s

We are calculating the time taken by the slower train to pass the driver of the faster one.

Hence, distance = length of the slower train = 500 metre
Time = 500/( 125/6)
= 24 seconds

Question: A train 300 meters long passes a pole in 20 seconds. How long will it take to pass a platform that is 500 meters long?

Solution:
Train's speed = Distance/Time
= 300/20 = 15m/s

Total distance to pass the platform,
= Length of train + Length of platform 
= 300 + 500
= 800m

∴ Required time = Distance/Speed
= 800/15
= 53.33 seconds

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x-1) hours = 25(x-1) km.
20x+25(x-1) = 110
45x = 135
So, x = 3.
So, they meet at 10 a.m.

Question: A train 400 meters long passes a pole in 16 seconds. How long will it take to pass a platform that is 800 meters long?

Solution:
Train's speed = Distance/Time
= 400/16 = 25m/s

Total distance to pass the platform,
= Length of train + Length of platform
= 400 + 800
= 1200 meters

∴ Required time to pass platform = Distance/Speed
= 1200/25
= 48 seconds

Question: The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 5.5 m away from the wall. The length of the ladder is-

Solution: 

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 5.5 m
AC/BC = cos⁡60^∘= 1/2
⇒ BC = 2 × AC = 2 × 5.5 = 11 m

Let the speed of the slower train = X m/sec
Then, the speed of the faster train will be = 2X m/sec

Relative Speed = X + 2X = 3X m/sec
Relative Speed is also = Sum of the length of the trains/Time taken to cross each other
= (200 + 200)/10
= 400/10
= 40.

So, 3X = 40
X = 40/3
X (speed) in Km/hr = (40/3) × (18/5)
= 720/15
= 48 km/hr.

Question: A boat travels 18 km downstream in 45 minutes. If the speed of the stream is 5 km/h, what is the speed of the boat in still water?

Solution:
স্রোতের অনুকূলে 45 মিনিটে যায় 18 কিমি
স্রোতের অনুকূলে 1 মিনিটে যায় 18/45 কিমি
স্রোতের অনুকূলে 1 ঘণ্টা বা 60 মিনিটে যায় (18 × 60)/45 কিমি
= 24 কিমি

∴ স্রোতের অনুকূলে বেগ = 24 কিমি/ঘণ্টা

দেওয়া আছে,
স্রোতের বেগ = 5 কিমি/ঘণ্টা।

∴ স্থির পানিতে নৌকার বেগ = স্রোতের অনুকূলে বেগ - স্রোতের বেগ
= 24 - 5 = 19 কিমি/ঘণ্টা।

Relative speed = (60+40) km/hr
= 100×(5/18) m/sec
= 250/9 m/sec.
Distance covered in crossing each other
= (140+160)m= 300m
Required time
= 300×(9/250) sec
= 54/5 sec
= 10.8 sec

Let the distance be D km.
∴ Downstream Speed = D/4 km/hr
And Upstream Speed = D/5 km/hr
Given, Speed of current = 2 km/hr
Speed of the current = 1/2 ×(Downstream Speed - Upstream Speed)
2 = 1/2 ×(D/4 - D/5)
D = 80 km

Question: A man takes 6 hours to walk to a place and ride back to the starting point. If he were to walk both ways, it would take him 9 hours. How much time would he take to ride both ways?

Solution:
Time taken in walking both ways = 9 hours ................(i)
Time taken in walking one way and riding back = 6 hours ........(ii)

By the equation (ii) × 2 - (i), we have:
Time taken by the man in riding both ways
= (6 × 2) - 9
= 12 - 9
= 3 hours

∴ The time taken by him to ride both ways is 3 hours.

Let speed upstream = x
Then,
Speed downstream = 2 x
Speed in still water = (2 x + x) /2 = 3x/2
Speed of the stream = (2 x − x)/2 = x/2
Speed in still water : Speed of the stream
= 3x/2 : x/2
= 3 : 1

Question: A jogger running at 9 kmph alongside a railway track is 240 meters ahead of the engine of a 120 meters long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

Solution: 
Relative Speed = (45 - 9) kmph
= 36 kmph
= 36 × (5/18)
= 10 m/s

The train must cover the initial distance separating it from the jogger and its own length to fully pass him.
∴ Total Distance = (240 + 120) meters
= 360 meters

Now,
Time = Distance/Speed
= 360/10 seconds
= 36 seconds

∴ The train will take 36 seconds to completely pass the jogger.

Question: Two boats are travelling towards each other at speeds of 46 km/h and 62 km/h, respectively. What is the distance between the two boats half a second before they collide?

Solution: 
Relative speed = (46 + 62) km/h
= 108 km/h
= [108 × (5/18)] m/s
= 30 m/s

We know,
Distance = Relative speed × Time
= [30 × (1/2)] m
= 15 m 

∴ Distance between the two boats half a second before collision = 15 meters 

Question: A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is-

Solution:

We can write three-quarters of a kilometre as 750 metres, 11(1/4) minutes as 675 seconds and 7(1/2) minutes as 450 seconds.

Rate upstream = (750/675) m/sec = 10/9 m/sec
Rate downstream = (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2) (10/9 + 5/3) m/sec
= 25/18 m/sec
= (25/18 × 18/5) km/hr
= 5km/hr

In still water, we know that
speed of boat (x) = (1/2) × (Downstream speed + upstream speed)
With the given parameters like S_D = 40 km/hr & S_U = 22 km/hr, on substituting these values in above equation,
we obtain,
x = (1/2) × [S_D + S_U]
⇒ x = (1/2) × (40 + 22)
⇒ x = 31 km/hr.

Question: P runs three times faster than Q, and Q runs twice as fast as R. If R covers a certain distance in 120 minutes, how long will it take P to cover the same distance?

Solution:
Let,
Speed of R = x
Then speed of Q = 2x
Speed of P = 2x × 3 = 6x

Ratio of speeds P : Q : R = 6x : 2x : x = 6 : 2 : 1

Since time is inversely proportional to speed,
Ratio of time taken P : Q : R = 1/6 : 1/2 : 1 = 1 : 3 : 6
Given that,
R covers a certain distance in 120 minutes.
So,
6 units = 120 minutes
1 unit = (120 ÷ 6) minutes = 20 minutes

∴ P will cover the same distance in 20 minutes.

Question: In a 500 m race, the speeds of two runners, A and B are in the ratio 5 : 6. If A is given a start of 100m, by how many meters does A win the race?

Solution:
Total race length = 500 meters.
A is given a start of 100 meters, so A runs 500 - 100 = 400 meters.

Speed ratio A : B = 5 : 6.

Let, B runs = X meter

Therefore,
400/X = 5/6
⇒ X = (6 × 400)/5
∴ X = 480m

Remaining distance for B = 500 - 480 = 20 meters.
Therefore, A wins by 20 meters.

Question: A fighter jet covers a certain distance at a speed of 1200 km/h in 5 hours. What speed must it maintain to cover the same distance in 250 minutes?

Solution:
Total distance = Speed × Time
= (1200 × 5) km
= 6000 km

Given time = 250 minutes = (250/60) hours= 25/6 hours

∴ Required speed = Distance/Time
= {6000/(25/6)} km/h
= {6000 × (6/25)} km/h
= (240 × 6) km/h
= 1440 km/h

Question: A man can row upstream at 10 kmph and downstream at 20 kmph. Find the man's rate in still water and the rate of the stream.

Solution:
If a is rate downstream and b is rate upstream 
Rate in still water = (a + b)/2 
Rate of current = (a - b)/2 

Rate in still water = (20 + 10)/2 = 15 kmph 
Rate of current = (20 - 10)/2 = 5 kmph 

Question: A boat while downstream in a river covered a distance of 50 miles at an average speed of 60 miles per hour. While returning, because of the water resistance, it took 1 hour 15 minutes to cover the same distance. What was the average speed during the whole journey?

Solution: 
Time taken to cover 50 miles downstream = (50/60)h
= 5/6 h
Time taken to cover 50 miles upstream = 1h 15m
= 5/4 h

Total time taken to cover 100 miles = (5/6) + (5/4)
= 25/12 h

∴ Average speed  = 100/(25/12)
= (100 × 12)/25
= 48 mph

Question: In one hour, a boat goes 11 km/hr along the stream and 7 km/hr against the stream. The speed of the boat in still water (in km/hr) is-

Solution:
Speed in still water = (11 + 7)/2 kmph
= 18/2 kmph
= 9 kmph.

Let, distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min late.
Time taken at 4 kmph : x/4 = 30 min earlier
Difference between time taken : 30 - (-20) = 50 mins = 50/60 hours.
x/3 - x/4 = 50/60
x/12 = 5/6
x = 10 km.

If Rashed is walking 5/6 of his usual speed that means he is taking 6/5 of using time.

According to the question,
6/5 of usual time - usual time = 10 mins
1/5 of usual time = 10 mins
Usual time = 50 mins.

Length of bridge = Distance travelled by the person in 4 min
= speed X time
Speed = 54 Km/hr = 54×5/18 = 3 × 5 = 15 m/s.
Time =4 min = 4 × 60 = 240 s
Required length = 15 × 240 = 3600 m.

Question: Two trains of equal lengths takes 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 miters, in what time ( in seconds) will they cross each other traveling in opposite direction?

Solution:
Speed of the first train = (120/10) m/sec = 12 m/sec

And,
Speed of the second train = (120/15) m/sec = 8 m/sec

∴ Relative speed = (12 + 8)m/sec = 20 m/sec

∴ Required time = (120 + 120)/20 sec
= 240/20
= 12 seconds

Question: A train 110 metres long is running at a speed of 60 km/h. In what time will it pass a man who is running at 6 km/h in the direction opposite to that in which the train is going?

Solution:
Given that,
Length of train = 110 meters
Speed of train = 60 km/h
Speed of man = 6 km/h (in opposite direction)

Now, Since the train and the man are moving in opposite directions, their relative speed is the sum of their individual speeds.
 S_relative = S_train + S_man = (60 + 6)km/h = 66 × (5/18) = 55/3 m/s

Than, 
The train passes the man when it covers its own length (110 m) at relative speed.

∴ Time = Distance/Relative speed
= 110/(55/3)
= 110 × (3/55)
= 2 × 3
= 6 seconds

∴ The train will pass the man in 6 seconds.

Question: Two trains 240 metres and 270 metres in length are running towards each other on parallel lines, one at the rate of 60 kmph and another at 48 kmph. How much time will they take to cross each other? 

Solution:
Given that,
Length of first train = 240 m
Length of second train = 270 m
Speed of first train = 60 km/h
Speed of second train = 48 km/h

∴ Relative speed = 60 + 48 = 108 km/h
= 108 × (5/18) m/s
= 30 m/s

And total distance to be covered to completely cross each other,
= Sum of lengths of both trains
= 240 m + 270 m
= 510 m

We know, 
Time taken = Distance/Relative speed
= 510m/30 m/s
= 17 seconds

So the two trains will take 17 seconds to cross each other.

Man walks 20 km in → 5 hours

That means it will take more time to walk 32 km.
This is the case of direct proportion.

20/5 = 32/x
x = 32/4
= 8

Question: A boat sails m miles upstream at r miles/hr. If the speed of the stream is  s miles/hr, how long will it take the boat to return to its starting point?

Solution:
মনেকরি
নৌকার গতিবেগ = x কিমি/ঘণ্টা
স্রোতের গতিবেগ = s কিমি/ঘণ্টা

এখানে
x - s = r
x = r + s

আবার
স্রোতের অনুকূলে বেগ =  x + s কিমি/ঘণ্টা
= r + s + s কিমি/ঘণ্টা
= r + 2s

স্রোতের অনুকূলে ফিরে আসতে সময় লাগবে = দূরত্ব/বেগ
= m/(r + 2s) 

Question: The speeds of three cars are the ratio 2 : 3 : 4. The ratio of the taken by these cars to travel the same distance is-

Solution:
Given that,
Speed ratio of three cars,
v₁ : v₂ : v₃ = 2 : 3 : 4 
Let, 
v₁ = 2k, v₂ = 3k, v₃​ = 4k (for some constant k)

We know,
Time = distance​/Speed
∴ t₁​ = d/2k​, t₂ ​= d/3k​, t₃ ​= d/4k

∴ Ratio of time = t₁ : t₂ : t₃ = d/2k : d/3k : d/4k 
= 1/2 : 1/3 : 1/4    ; [Cancel d and k (since d,k ≠ 0)]​
= 12/2 : 12/3 : 12/4.  ; [LCM of 2, 3, 4 = 12]
= 6 : 4 : 3

So the ratio of the time taken is 6 : 4 : 3

Let speed upstream be x km/hr
Then, speed downstream = 3x km/hr.

Speed in still water = 1/2 (3x + x) kmph = 2x km/hr.

∴ 2x = 28/3
x = 14/3 km/hr;
Speed downstream = 14 km/hr

Hence, speed of the current
1/2 {14 - (14/3)} km/hr
= 14/3 km/hr
= 4(2/3) km/hr.

Let the length of the train be x
∴ x/20 = x + 250/45
⇒ 5x = 1000
⇒ x = 200 m

When B runs 25m, A runs 45/2m
When B runs 1000m,
A runs = (45/2 × 1/25 × 1000)m
= 900m
∴ B beats A by 100 m

Question: A car moves from Dhaka to Cumilla at the original speed at 60 kmph and reached Cumilla in 20 minutes late. If the car increased speed by one fourth of his original speed reached Cumilla on time, find the distance between Dhaka and Cumilla?

Solution:
Let distance = d km.
Original speed = 60 km/h. So time taken = d/60 hours.

But it is 20 minutes late. That means the scheduled time T (in hours) is such that d/60 = T + (20/60) = T + (1/3).

And
When speed is increased by one fourth. So new speed = 60 + (1/4) × 60 = 60 + 15 = 75 km/h.
Then time taken = d/75. And this equals T (on time).

So we have,
d/60 = T + 1(/3) .........(1)
d/75 = T ..........(2)

Now, subtract the second equation from the first,
(d/60) - (d/75) = 1/3
⇒ (5d - 4d)/300 = 1/3
⇒ d = 300/3
∴ d = 100 km

∴ Distance between Dhaka and Cumilla = 100 km

Question: The speed of a boat down the stream is 125% of the speed in still water. If the boat takes 30 minutes to cover 20 km in still water, then how much time (in hours) will it take to cover 15 km upstream?

Solution: 
Given that, 
Speed downstream = 125% of speed in still water
Distance in still water = 20 km, time = 30 min = 0.5 hr
Distance to travel upstream = 15 km

Speed of boat in still water = Distance/Time = 20/0.5 = 40 km/h
And,
Downstream speed = 125% of still water speed.
∴ Downstream speed = 1.25 × 40 = 50 km/h 

We know,
Downstream speed = Boat speed in still water + Current speed 
⇒ 50 = 40 + Current speed 
⇒  Current speed = 50 - 40 = 10 km/h 

And upstream speed = Boat speed in still water - Current speed = 40 - 10 = 30 km/h 

∴ Time to cover 15 km upstream = Distance/Speed ​= 15/30 = 1/2 = 0.5 hours

So the time required for the boat to cover 15 km upstream is 0.5 hours. 

Question: A wheel of an engine of 450 cm in circumference makes 20 revolutions in 6 seconds. What is the speed of the wheel in km/h?

Solution:
Total distance = (450 × 20) cm
= 9000 cm
= 9000/100 m
= 90 m

We know,
Speed = (Total distance ÷ Time)
= (90 ÷ 6) m/sec
= 15 m/sec
= (15 × 18/5) km/h
= 54 km/h

Question: In a 500-meter race, Q starts 50 meters ahead of P, yet P defeats Q by a margin of 25 meters. What distance did Q cover when P reached the finish line?

Solution:
Total distance Q needed to cover = 500 - 50 = 450 meters
Distance covered by P = 500 meters
But P defeats Q by 25 meters

∴ Distance covered by Q when P reaches the finish line = 450 - 25 = 425 meters

∴  Q covered a distance of 425 meters when P reached the finish line.

অতিক্রান্ত সময়, t = (d₁ + d₂)/(v₁ + v₂)
= (105 + 90)/{(45 + 72) × 5}/18
= 195/(117 × 5)/18
= 195 × 18/(117 × 5)
= 6 seconds

Question: A boat goes 10 km upstream in 50 minutes, and the speed of the stream is 3 kmph. Find the speed of the boat in still water (in km/h). 

Solution: 
Let the speed of the boat in still water = x km/h
Speed of the stream = 3 km/h (given)
Upstream speed = x - 3 km/h
Distance upstream = 10 km
Time upstream = 50 minutes = 50/60 hours = 5/6 hours

We know,
Speed = Distance/Time
Upstream speed = 10 /(5/6) = 10 × (6/5) = 12 km/h
So,
⇒ x - 3 = 12
⇒ x = 12 + 3
∴ x = 15 km/h

The speed of the boat in still water is 15 km/h.

Question: A man completes a journey in 8 hours. He travels the first half of the journey at the rate of 40 km/hr and the second half at the rate of 60 km/hr. Find the total journey in km.

Solution:
ধরা যাক, মোট যাত্রার দূরত্ব হলো D কিমি।
তাহলে, যাত্রার প্রথম অর্ধেকের দূরত্ব হবে D/2 কিমি
এবং দ্বিতীয় অর্ধেকের দূরত্বও হবে D/2 কিমি।

প্রথম অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (D/2)/40 ঘন্টা
= D/80 ঘন্টা

দ্বিতীয় অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (D/2)/60 ঘন্টা
= D/120 ঘন্টা

প্রশ্নমতে,
D/80 + D/120 = 8
⇒ (3D + 2D)/240 = 8
⇒ 5D/240 = 8
⇒ 5D = 8 × 240
⇒ 5D = 1920
⇒ D = 1920/5
⇒ D = 384 কিমি

∴ মোট যাত্রার দূরত্ব 384 কিলোমিটার।