Time and Speed - Train and Boat

Question: A train takes 50 seconds to cross a platform 300 meters long and 35 seconds to cross another platform 150 meters long. Find the length of the train.

Solution:
Let the length of the train = x meters

Then,
For the first platform, the distance covered by the train = (x + 300) meters
And,
For the second platform, the distance covered by the train = (x + 150) meters

According to the question,
(x + 300)/50 = (x + 150)/35
⇒ 50(x + 150) = 35(x + 300)
⇒ 50x + 7500 = 35x + 10500
⇒ 50x - 35x = 10500 - 7500
⇒ 15x = 3000
⇒ x = 3000/15 
⇒ x = 200

∴ The length of the train is 200 meters

Speed upstream = 3/(20/60)
= 9 km/hr.
Speed downstream = 3/(18/60)
= 10 km/hr.
Rate of current = (10 - 9)/2
= 1/2 km/hr.

Question: A man rowed 4 miles upstream in 2 hours. If the river current is 1 mile per hour, how long will it take him to return downstream?

Solution:
Let,
the velocity of the boat in still water = x mph
and the velocity of the stream = y mph = 1 mph

In upstream, he covers 4 miles in 2 hours.
∴ Upstream speed = 4 ÷ 2 = 2 mph

ATQ,
x - y = 2
⇒ x - 1 = 2
⇒ x = 3

So, downstream speed = x + y = 3 + 1 = 4 mph

Time to row 4 miles downstream = 4 ÷ 4 = 1 hour

∴ The return trip will take 1 hour.

Question: A cyclist completes a journey in 8 hours. He travels the first half of the journey at the rate of 15 km/hr and the second half at the rate of 25 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)/15 ঘন্টা
= D/30 ঘন্টা

দ্বিতীয় অর্ধেক যাত্রায়, সময় = দূরত্ব/গতিবেগ
= (D/2)/25 ঘন্টা
= D/50 ঘন্টা

প্রশ্নমতে,
 D/30 + D/50 = 8
⇒ (5D + 3D)/150 = 8
⇒ 5D + 3D = 8 × 150
⇒ 8D = 1200
⇒ D = 1200/8
⇒ D = 150 কিমি

সুতরাং, মোট যাত্রার দূরত্ব হলো 150 কিমি।

The man starts with 7 kmph
Distance covered in first 1 hour = 7 km
He increases his speed every hour by 1 km.
Speed in 2nd hour = 8 km/hr
Distance covered in 2nd hour = 8 km
Remaining distance = 19.5 - (7 + 8)
19.5 - 15
= 4.5 km
Speed in the third hour = 9 km/hr
Time taken to cover 4.5 km at 9 km/hr
= 4.5/9
= 1/2 hour.
Therefore,
Total time = 1 + 1 + (1/2) hours = 2 (1/2) hours.
Hence, the answer is 2(1/2) hours.

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

Let the speed of the slower train be x m/sec.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x+2x) m/sec = 3x m/sec.
So,(100 + 100)/8 = 3x
⇒ 24x = 200
⇒ x = 25/3
So, speed of the faster train =50/3 m/sec
= (50/3) x (18/5) km/hr
= 60 km/hr.

Given that,
The speed of a boat in still water is 20 km/hr
The speed of the stream is 4 km/hr.
Then,
The speed downstream = (20 + 4) km/hr
= 24 km/hr
The distance travelled in half an hour = 24 x 1/2
= 12 km
Hence the distance travelled by the boat is 12 km.

Let rate of stream be y km/hr, distance be d km
time upstream = 2 (time downstream)
⇒ d/(7.5 − y) = 2d/(7.5 + y)
⇒ 2( 7.5 − y ) = 7.5 + y
⇒ 15 − 2 y = 7.5 + y
⇒ 3 y = 7.5
⇒ y = 2.5

Question: A train having length 150 m passes a platform of 550 m length. The time taken for it is 56 seconds. In how much time will this train take to pass a platform of 250 m length?

Solution:
We know,
When a train passes a platform, the total distance to be covered is,
Length of train + Length of platform

Case 1:
Distance covered = 150 m + 550 m = 700 m
Time taken = 56 seconds
∴ Speed of the train = Distance/Time
= 700/56
= 12.5 m/s

Case 2: 
Length of new platform = 250 m
∴ Total distance to be covered now = 150 + 250 = 400 m
Time taken to pass the new platform = Distance/Speed
= 400m/12.5 m/s
= 32 seconds

So the train will take 32 seconds to pass the platform of 250 m length.

Relative speed = (45 + 30) km/hr.
= 75 km/hr.
= 75 × (5/18) km/hr.
= (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 meter.
Time = 500/(125/6)
= 500 × (6/125)
= 24 seconds.

Distance covered = (108 + 112) = 220 meter.
Time taken = 6 seconds.
Relative speed = 220/6 = 110/3 m/s.
= (110/3) × (18/5) km/hr
= 132 km/hr.

ATQ, 50 + Speed of second train = 132 km/hr.
⇒ Speed of second train = (132 - 50) = 82 km/hr.