Time and Speed - Train and Boat

Question: Sonar Bangla express, normally reaches its destination at 72 km/h in 6 hours. Find the speed at which it must travel to reduce the travel time by 2 hours?

Solution:
Distance to be covered = Speed × Time
= 72 × 6
= 432 km

Reduced time = 6 - 2 = 4 hours

∴ Required speed = 432/4
= 108 km/h

∴ The train must travel at 108 km/h to reduce the time by 2 hours.

Question: Without stoppages, the bus goes 72 kmph, and with stoppages, 60 kmph. Calculate the bus’s stopping time per hour. 

Solution:
Including stoppages, it covers = (72 - 60) kmph
= 12 kmph less

Time taken to cover 12 km = {(12/72) × 60} minutes
= 10 minutes

Question: A man rows at 6 km/hr in still water and 4.5 km/hr against current. His rate along the current is:

Solution:
Speed in still water is 6 km/hr
Speed of the current is x km/hr

against current,
6 - x = 4.5
⇒ x = 6 - 4.5
∴ x = 1.5

along the current,
6 + x
= 6 + 1.5
= 7.5 km/hr

Question: Walking 3/4 of his normal speed, Rafi is 12 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office-

Solution:
Let s be Rafi's normal speed,  t be his usual time, and t' be his new time.

Since the distance to the office is constant, d = s × t
When Rafi walks at his normal speed his new time is expressed as, d = (3/4)s × t'

Again, since the distance is the same,
st = (3/4)s × t'
⇒ t' = (4/3)t 

ATQ,
t' - t = 12
⇒ (4/3)t - t = 12
⇒ (4t - 3t)/3 = 12
⇒ t/3 = 12
∴ t = 36 min

Time taken,
7.25 P.M. - 4.10 P.M. = 3 hr. 15 min.
= 3 × (15/60) hrs.
= 3(1/4) hrs.
= (13/4) hrs.
∴ Required distance = {40 × (13/4)} km.
= 130 km.

Question: With an average speed of 50 km/hr, a train reaches its destination in time. If it goes with an average speed of 40 km/hr, it is late by 30 min. The total journey is-

Solution:
Difference between timings = 30 min = 30/60 hr = 1/2 hr.
Let the length of the journey be x km.

Then,
(x/40) - (x/50) = 1/2
(5x - 4x)/200 = 1/2
x/200 = 1/2
x = 200/2
∴ x = 100 km.

∴ The total journey is 100 km.

Let the speed of the first train be x km/hr.
Then,
Sum of lengths of trains = (66 + 88)m = 154 m.

Relative speed of two trains = (x - 30) km/hr
= {(x - 30) × (5/18)} m/s

∴ 154/(x - 30) × (5/18) = 0.168 × 60
⇒ 5(x - 30) = (154 × 18)/10.08
⇒ 5(x - 30) = 275
⇒ x - 30 = 55
⇒ x = 85 km/hr.

Question: A boy rides his bicycle 10 km at an average speed of 10 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is-

Solution:
Given that,
Distance₁ = 10 km, Speed₁ = 10 km/hr
Distance₂ = 12 km, Speed₂ = 10 km/hr

∴ Total Distance = 10 + 12 = 22 km

And,
Time₁ = 10 ÷ 10 = 1 hr
Time₂ = 12 ÷ 10 = 1.2 hr

∴ Total Time = 1 + 1.2 = 2.2 hr

We know,
Average Speed = Total Distance ÷ Total Time
= 22/2.2 = 10 km/hr

∴ His average speed for the entire trip is 10 km/hr.

Question: Two trains A and B are moving in the same direction. A has speed of 8 km/h and B has speed of 13 km/h. What is relative speed of B with respect to A?

Solution: 
Given that,
Speed of train A = 8 km/h
Speed of train B = 13 km/h

Since both are moving in the same direction, and B is faster.
∴ Relative speed of B with respect to A = Speed of B - Speed of A
= 13 km/h - 8 km/h
= 5 km/h

So the relative speed of B with respect to A is 5 km/h.

Let length of each train = x metre
Total distance covered while passing the slower train = (x + x) = 2x metre
Relative speed = (46 − 36)
= 10 km/hr
= 10 × 5/18
= 50/18 m/s

Time = 36 seconds
⇒ 2 x/36 = 50/18
⇒ x = 50

Velocity of the river = [100 √{(1/4²) - (1/5²)}] m/min
= [100 √{(1/16) - (1/25)}] m/min
= 100{√(9/400)} m/min
= 100 × (3/20) m/min
= 15 m/min.

Required ratio:
= (65×8):(70×4)
= 520:280
= 13:7

Question: Maruf is travelling on his cycle and he calculated to reach point A at 3 p.m. if he travels at 10 kmph, he will reach there at 1 p.m. if he travels at 15 kmph. At what speed must he travel to reach A at 2 p.m.

Solution: 
Let the distance travelled by x km.

ATQ, 
x/10 - x/15 = 2 
⇒ 3x - 2x = 60 
∴ x = 60

Time taken to travel 60 km at 10km/h = 60/10 hours
= 6 hours

So, Maruf started 6 hours before 3 P.M. i.e., at 9 Α.Μ.

∴ Required speed = 60/5 kmph. 
= 12 kmph.

Total distance travelled = (39 + 25)
= 64 km
Total time taken = (45 + 35)
= 80 min.
= (80/60) hr.
= (4/3) hr.
∴ Average speed = {64 × (3/4)} km/hr
= 48 km/hr.
Hence, the average speed of the car is 48 km/hr.

Train’s speed = 240/24 = 10 m/s
The train has to cover = (240 + 650) = 890 m.
∴ Required time = 890/10 = 89 seconds

Question: A man travelled a distance of 1260 km in 17 hours. He travelled partly by car at a speed of 40 km/h, and partly by train at a speed of 80 km/h. What is the distance traveled by the train?

Solution: 
Given,
 Total distance = 1260 km
Total time = 17 hours
Speed of car = 40 km/h
Speed of train = 80 km/h

Let,
Distance travelled by car = x km
Distance travelled by train = (1260 - x) km

ATQ,
(x/40) + [(1260 - x)/80] = 17
⇒ 2x + 1260 - x = 17 × 80
⇒ x = 1360 - 1260
⇒ x = 100

∴ Distance traveled by the train = (1260 - 100) km
= 1160 km

Question: Two trains running in opposite directions cross a man standing on the platform in 17 seconds and 7 seconds respectively and they cross each other in 13 seconds. The ratio of their speeds is-

Solution: 
Let, the speed of the trains are x m/sec and y m/sec respectively

We know, Distance/Length = Speed × Time
∴ Length of first train = 17x meter
And, length of second train = 7y meter

Now, Total time = Total length/Total speed
(17x + 7y)/(x + y) = 13
⇒ 17x + 7y = 13x + 13y 
⇒ 17x - 13x = 13y - 7y
⇒ 4x = 6y
⇒ x/y = 3/2
∴ x ∶ y = 3 ∶ 2

Question: A train running at the speed of 60 kmph crosses a 200 m long platform in 27 seconds. What is the length of the train?

Solution: 
Given that,
Time = 27 sec
∴ Speed = (60 × 5/18)m/sec = 50/3 m/sec

Let the length of the train be x metres.
Then,
(x + 200)/27 = 50/3
⇒ x + 200 = (50/3) × 27 
⇒ x + 200 = 450
⇒ x = 450 - 200
∴ x = 250 metres

So the length of the train is 250 meters.

15min = 1/4hrs
1 hr → 5 kms
1/4hr → 5/4 kms
So, length of the bridge=5/4 km = 1250 metres

By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minutes drop in his time.)
Hence, his time originally must have been 30 minutes.
Thus required distance = 20 kmph × 0.5 hours = 10 km.

Question: Two trains start from station P and Q and travel towards each other at speeds of 60 km/hr and 80 km/hr respectively. At the time of their meeting, the second train has traveled 80 km more than the first. The distance between P and Q is:

সমাধান:
দ্বিতীয় ট্রেনের গতিবেগ বেশি = (80 - 60) কিমি/ঘন্টা
= 20 কিমি/ঘন্টা।

দ্বিতীয় ট্রেনটি প্রথম ট্রেনের চেয়ে 80 কিমি বেশি ভ্রমণ করেছে।
এই বাড়তি দূরত্ব অতিক্রম করার সময় = দূরত্ব/গতিবেগ
= 80 কিমি/20 কিমি/ঘন্টা
= 4 ঘন্টা

যেহেতু ট্রেন দুটি একই সময়ে যাত্রা শুরু করেছিল, তাই তাদের মিলিত হতে 4 ঘন্টা সময় লেগেছে।

ট্রেন দুটির আপেক্ষিক গতিবেগ (Relative speed) = (60 + 80) কিমি/ঘন্টা
= 140 কিমি/ঘন্টা।

সুতরাং, মোট অতিক্রান্ত দূরত্ব = আপেক্ষিক গতিবেগ × সময়
= 140 কিমি/ঘন্টা × 4 ঘন্টা
= 560 কিমি

সুতরাং, P এবং Q স্টেশনের মধ্যবর্তী দূরত্ব হলো 560 কিমি।

Question: A boat running upstream takes 12 hours to cover a certain distance, while it takes 8 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

Solution:
ধরি,
স্থির পানিতে নৌকার গতি = b কিমি/ঘণ্টা
স্রোতের গতি = c কিমি/ঘণ্টা
দূরত্ব = d কিমি

স্রোতের প্রতিকূলে নৌকার গতিবেগ = (b - c) কিমি/ঘণ্টা
∴ d/(b - c) = 12
⇒ d = 12(b - c) 

স্রোতের অনুকূলে নৌকার গতিবেগ = (b + c) কিমি/ঘণ্টা
∴ d/(b + c) = 8
⇒ d = 8(b + c) 

এখন, উভয় ক্ষেত্রে দূরত্ব সমান হওয়ায়,
12(b - c) = 8(b + c) 
⇒ 3(b - c) = 2(b + c)
⇒ 3b - 3c = 2b + 2c
⇒ b = 5c
⇒ b/c = 5/1

∴ b : c = 5 : 1

Question: A train of 180 m long is moving at 72 km/h. The time taken by the train to cross a tunnel of 420 m long is-

Solution:
মোট অতিক্রান্ত দূরত্ব = ট্রেনের দৈর্ঘ্য + টানেলের দৈর্ঘ্য
= (180 + 420) মিটার
= 600 মিটার

ট্রেনের গতিবেগ = 72 কিমি/ঘন্টা
= (72 × 1000) মিটার/3600 সেকেন্ড
= 20 মিটার/সেকেন্ড

সময় = দূরত্ব ÷ গতিবেগ
= 600 মিটার ÷ 20 মিটার/সেকেন্ড
= 30 সেকেন্ড

সুতরাং, টানেলটি অতিক্রম করতে ট্রেনটির 30 সেকেন্ড সময় লাগবে।

Since the train is late, the allotted time = Y - 2.
D = T × S
X = (Y - 2) × S
or, S = X/(Y - 2)

Distance covered in 26 seconds.
= 26 × 72 × (5/18)
= 520 meter.
Length of the train = (520 - 250) meter
= 270 meter.

Speed of the stream = 1
Motor boat speed in still water be = x kmph
Downstream = x + 1 kmph
Up Stream = x - 1 kmph
[35/(x + 1)] + [35/(x - 1)] = 12
x = 6 kmph

(d/80 + d/40) - (d/60 + d/60) = 2
(3d + 6d - 4d - 4d)/240 = 2
d = 480 km

Speed upstream = 2/2 = 1 km/hr
Speed downstream = 1/(20/60) = 3 km/hr
Speed in still water = (1/2)(3+1) = 2 km/hr
Time taken to travel 5 km in still water = 5/2
= 2(1/2)
= 2 hr 30 min

Question: A train passes a station platform in 42 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 72 km/hr, what is the length of the platform?

Solution:
ট্রেনটি 3600 সেকেন্ডে অতিক্রম করে 72000 মিটার 
ট্রেনটি 1 সেকেন্ডে অতিক্রম করে 72000/3600 মিটার 
ট্রেনটি 20 সেকেন্ডে অতিক্রম করে (72000 × 20)/3600 মিটার 
= 400 মিটার 

ট্রেনের দৈর্ঘ্য = 400 মিটার 
আবার 
ট্রেনটি 3600 সেকেন্ডে অতিক্রম করে 72000 মিটার 
ট্রেনটি 1 সেকেন্ডে অতিক্রম করে 72000/3600 মিটার 
ট্রেনটি 42 সেকেন্ডে অতিক্রম করে (72000 × 42)/3600 মিটার 
= 840 মিটার 

প্লাটফর্মের দৈর্ঘ্য  = (840 - 400) = 440 মিটার 

Let,
the speed of the stream be = u kmph
the speed of the boat be = v kmph
speed upstream will be = v - u kmph
speed downstream will be = v + u kmph

30 km upstream in time duration = 30 / (v - u) hrs
44 km downstream in time duration = 44 / (v + u) hrs
44/(v + u) + 30/(v - u)
= 10 hrs .........(i)

Similarly,
40 /(v - u) + 55 / (v + u) = 13 hrs Multiply with 3/4:
30/(v - u) + 165 / 4(v + u) = 39/4 .......(ii)

Now (i) - (ii)
⇒ [44 - 165/4] / (v + u) = 10 - 39/4 = 1/4
⇒ v + u = 11 ..........(iii)
Substitute this in (1) to get: 44/11 + 30/(v - u) = 10
⇒ v - u = 30/6 = 5.........(iv)

Solving (iii) and (iv) ,
we get :
v = 8 kmph and
u = 3 kmph

Now, Let the distance to the point be x km.
When swimming downstream, the swimmer's speed = (5 + 1) = 6 km/h.
When swimming upstream, the swimmer's speed = (5 - 1) = 4 km/h

Now,
T₁ + T₂ = T
⇒ (x/6) + (x/4) = 5/4
⇒ (2x + 3x)/12 = 5/4
⇒ 5x = 15
⇒ x = 15/5
∴ x = 3 km.

∴ The place is 3 km away from the starting point.

Question: An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 5/3 hours, it must travel at a speed of-

Solution: 
Given that,
Speed = 240 kmph 
Time = 5 hours 

We know,
 Distance = Speed × Time 
= 240 × 5
∴ Distance = 1200 km

Again,
Given: 
Distance = 1200 km 
New Time = 5/3 hours 
 
 New Speed = Distance/New Time
= 1200/(5/3)
= (1200 × 3)/5
= 720 kmph
 
 ∴ The aeroplane must travel at a speed of 720 kmph.