Time and Speed - Train and Boat

Let after T hours they meet
Then, 3T+4T=17.5
T=2.5
Time = 10:00 am + 2.5 hour = 12:30 pm

বাইকারের যাত্রা শুরুর গতি = 60 কি.মি/ঘন্টা
প্রথম 2 ঘন্টায় যায় (60 × 2) = 120 কি.মি
পরের 2 ঘন্টায় যায় (63 × 2) = 126 কি.মি
আবার, পরের 2 ঘন্টায় যায় (66 × 2) = 132 কি.মি
এরকম 12 বার হবে।
সুতরাং, এটি একটি সমান্তর ধারা।

ধারাটি হবে, 120 + 126 + 132 + ------
এখানে ১ম পদ (a) = 120, সাধারণ অন্তর (d) = 6 ও পদসংখ্যা (n) = 12
∴ সমষ্টি (S) = (n/2) {2a + (n - 1)d}
= (12/2) { 2×120 + (12 - 1) × 6}
= 6 × 306
= 1836

সুতরাং, সঠিক উত্তর ঘ) none of these

15min = 1/4hrs
1 hr → 4 kms
1/4hr → 4/4 kms
So, length of the bridge= 1 km = 1000 metres

In the first minute the monkey climbs 1 meter.
In the second minute it slips 1/2 meter.
For every two minute it climbs 1/2 meter.

So Average speed = 1 meter/4 minutes
For 11 meters, time taken = 44 minutes.

For the last 1 meter jump add 1 minute.
So time taken = 45 minutes.

Question: Two trains are moving in opposite directions at speeds of 60 km/h and 90 km/h. Their lengths are 800 m and 700 m respectively. Find the time taken to cross each other.

Solution:
Relative speed = (60 + 90) km/h
= 150 × (5/18) m/sec
= 125/3 m/sec

Distance covered = (800 + 700) m
= 1500 m

Required time
= 1500 ÷ (125/3) sec
= (1500 × 3)/125 sec
= 36 sec

∴ The required time is 36 seconds.

প্রশ্ন: A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

সমাধান: 
Let,
The usual speed is x 
The distance is d 
The usual time is t min 

So, 
In usual speed, d = xt 

In 1/3 of usual speed, d = (x/3) × (t + 30) 

Now, We can say that,
xt = x(t + 30)/3 
⇒ t = (t + 30)/3
⇒ 3t = t + 30 
⇒ 2t = 30
∴ t = 15 

∴ The usual time to cover the same distance is 15 min.

Let the speed of the trains be 7 x and 8 x respectively.
Speed of second train = 400 /4 = 100 km/hr
⇒ 8 x = 100
⇒ x = 100/8 = 12.5
Speed of the first train = 7x = 7 × 12.5
= 87.5 km/hr

Question: The speed of three cars is in the ratio of 3 : 4 : 5. The ratio of the times taken by these cars to travel the same distance is-

Solution:
যেহেতু দূরত্ব একই থাকে, তাই গতিবেগ সময়ের সাথে ব্যস্তানুপাতিক হয়।
⇒ গতিবেগ ∝ (1/সময়)
⇒ s ∝ (1/t)

∴ সময়ের অনুপাত = 1/3 : 1/4 : 1/5
= (1/3 × 60) : (1/4 × 60) : (1/5 × 60)
= 20 : 15 : 12

Let the length of the train is x m. and its speed is v. m/s.

Distance = Speed × time [S = V × T]
x = v × 9 .........(i).
(x+700) = v × 30 ........(ii).

Dividing the eqn. (i) by (ii).

x/(x+700)= 3/10.
⇒ 10x=3x + 2100.
⇒ 7x=2100.
⇒ x= 2100/7.
⇒ x= 300. m.

putting x = 300 in eqn. (1).
300 = v × 9
⇒ v = 300/9
⇒ v = 100/3 m/s.

Let the train crosses a 800 m. long platform in t seconds.
(x + 800) = v × t .........(iii) [ S = V × T]
⇒ (300 + 800) = (100/3) × t. [putting x= 300. and v= 100/3.]
⇒ t = (1100×3)/100
⇒ t = 33 seconds.

AC = 40 × 4
= 160 km
BD = 35 × 4
= 140 km
BC = 200 - 160
= 40 km
AD = 200 - 140
= 60 km
∴ CD = 200 - BC - AD
= 200 - 40 - 60
= 100 km
Hence, 100 km apart will the two cars be after four hours of continuous travelling.

Question: A bus was supposed to travel 240 km at its normal speed. But because of heavy traffic, it had to slow down by 10 km/h and therefore reached the destination 2 hours later than the scheduled time. What was the bus’s original (normal) speed?

Solution:
Let, the original speed = x km/hr
Distance = 240 km

Time taken at original speed = 240 / x
Time taken at reduced speed = 240 / (x - 10)

According to the question:
240 / (x - 10) - 240 / x = 2

LCM: x(x - 10)
Now,
240x - 240(x - 10) = 2x(x - 10)
⇒ 240x - 240x + 2400 = 2x² - 20x
⇒ 2x² - 20x - 2400 = 0
⇒ x² - 10x - 1200 = 0

Solve quadratic: x² - 10x - 1200 = 0

Factors: (x - 40)(x + 30) = 0
x = 40 or -30 (speed cannot be negative)
So the speed is 40 km/hr.

Let the speed of A = x kmph and that of B = y kmph

According to the question,
(x × 6) + (y × 6) = 60
⇒ x + y = 10 --------- (i)
And,
(2x/3) × 5 + (2y × 5) = 60
⇒ 10x + 30y = 180
⇒ x + 3y = 18 ---------- (ii)
From equation (i) × 3 - (ii)
3x + 3y - x - 3y = 30 - 18
⇒ 2x = 12
Hence, x = 6 kmph.

Question: Kobita runs 5/2 times as fast as Babita. In a race, if Kobita gives a lead of 40 m to Babita, find the distance from the starting point where both of them will meet (correct up to two decimal places).

Solution:
Given that,
Kobita runs 5/2 times as fast as Babita
Kobita gives a lead of 40 m to Babita 

We know,
Distance = Speed × Time

Let the speed of Babita be = 2x
∴ Speed of Kobita = (5/2) × 2x = 5x

And,
Let the distance covered by Kobita be y meters
∴ Distance covered by Babita = (y - 40) meters

As time is constant, distance is directly proportional to speed,
2x/5x = (y - 40)/y
⇒ 2/5 = (y - 40)/y
⇒ 2y = 5y - 200
⇒ 3y = 200
⇒ y = 200/3
∴ y = 66.67 m

∴ The distance from the starting point where both of them will meet is 66.67 m.

Question: A biker travels at 60 km/h. If instead, he had traveled at 80 km/h for the same duration, he would have covered 100 km more. How far did he actually travel?

Solution:
Let, the actual distance travelled be x km.
Then,
x/60 = (x + 100)/80 
⇒ x/6 = (x + 100)/8
⇒ 6(x + 100) = 8x
⇒ 6x + 600 = 8x
⇒ 8x - 6x = 600
⇒ 2x = 600
⇒ x = 600/2
⇒ x = 300 km

We are given that two-thirds of the 6 km was covered at 4 km/hr i.e. 4 km distance was covered at 4 km/hr.
Time taken to cover 4 km = 4 km/4 km/hr = 1 hr = 60 minutes.
Time left = 84 – 60 = 24 minutes

Now, the man has to cover the remaining 2 km in 24 minutes or 24/60 = 0.4 hours
Speed required for remaining 2 km = 2 km/0.4 hr = 5 km/hr

Let the length of the train be x metres.
Then, length of the platform = 2x metres.
Speed of the train = 90× (5/18) m/sec
= 25m/sec
∴(x+2x)/25 = 36
⇒ 3x= 900
⇒ x= 300
Hence, length of platform
= 2x= (2×300)m= 600m

Question: A car travels at a speed that is 3/4th the speed of a bike. The bike covers 240 km in 4 hours. How much distance will the car cover in 30 minutes?

Solution:
Speed of the bike = distance/time
= 240/4 = 60 km/h

And speed of the car = 3/4 of the speed of the bike
= (3/4) × 60 = 45 km/h

∴ Time for the car = 30 minutes
= 30/60 = 1/2 hours

∴ Distance covered by the car = speed × time
= 45 × (1/2) = 22.5 km

So the car will cover 22.5 km in 30 minutes.

Question: A train takes 10 seconds to cross a pole and 25 seconds to cross a platform of length 180m. What is the length of the train?

Solution:
মনে করি, ট্রেনটির দৈর্ঘ্য L মিটার।

আমরা জানি, একটি খুঁটি (pole) অতিক্রম করার সময় ট্রেনটি কেবল তার নিজের দৈর্ঘ্য অতিক্রম করে।
সুতরাং, ট্রেনের গতিবেগ = L/10 মি./সে.  [গতিবেগ = দূরত্ব/সময়]

আবার, প্ল্যাটফর্ম অতিক্রম করার সময় ট্রেনটি (নিজের দৈর্ঘ্য + প্ল্যাটফর্মের দৈর্ঘ্য) অতিক্রম করে।
শর্তমতে, গতিবেগ = (L + 180)/25 মি./সে.

যেহেতু গতিবেগ একই, তাই,
L/10 = (L + 180)/25
⇒ 25L = 10(L + 180)
⇒ 25L = 10L + 1800
⇒ 25L - 10L = 1800
⇒ 15L = 1800
⇒ L = 1800/15
∴ L = 120

∴ ট্রেনটির দৈর্ঘ্য 120 মিটার।

Let the velocity or speed of the boat in still water is x km/hr.
And the Speed of the stream = 1km/hr

So, the speed of the boat along the stream = (x+1) km/hr.
The speed of the boat against the stream = (x-1) km/hr.

Note: time = Distance/Speed

So, [4/(x + 1)] + [4/(x - 1)] = 3 hrs.
⇒ [4 (x + 1 + x - 1)]/[(x + 1) (x - 1)] = 3
⇒ 8x = 3(x² - 1)
⇒ 8x = 3x² - 3
⇒ 3x² - 8x - 3=0
⇒ 3x² - 9x + x - 3 = 0
⇒ (x - 3) (3x + 1) = 0
Therefore x = 3 or, x = -1/3 (speed can't be -ve)

∴ Hence, the speed or velocity of the boat in still water is 3 km/hr.

Relative speed = 5.5 - 5
= 0.5 kmph (because they walk in the same direction)
Distance = 8.5 km
Time = Distance/Speed
= 8.5/0.5
= 17 hr.

Length of the train = 150 m
Speed of the man = 2 km/hr
Relative speed = 150/3 = 50 m/s
= 50 × 18/5
= 180 km/hr
Relative speed = Speed of train - Speed of the man (as both are moving in the same direction).
Therefore,
Speed of the train = Relative speed + Speed of the man
= 180 + 2
= 182 km/hr

Question: How many seconds will a 500 metre long train take to cross a man walking with a speed fo 3 km/ hr in the direction of the moving train if the speed of the train is 63 km/hr?

Solution:
Speed of the train =(63 - 3)km/hr
= 60km/hr
= (60 × 1000)/3600 m/sec
= 50/3

Time taken to pass the man = 500/(50/3) sec
= 500 × (3/50) sec
= 30 sec

Question: Ashik and Ruby run a race with their speed in the ratio of 5 : 3. They prefer to run on a circular track of circumference 2 km. What is the distance covered by Ashik when he passes Ruby for the sixth time?

Solution: 
Since the speeds of Ashik and Ruby are in the ratio 5 : 3 i.e., when Ashik covers 5 rounds, then Ruby covers 3 rounds
∴ Relative speed = 5 – 3 = 2 parts
 Now, the first time Ashik and Ruby meet, when Ashik completes (5/2 = 2.5) rounds, and Ruby completes 1/2 round.

∴ Ashik to pass Ruby for the sixth time, Ashik would have completed = (6 × 2.5) rounds
= 15 rounds

Since each round is 2 km,
Hence, the distance covered by Ashik = (15 × 2) km
= 30 km

We Know
Total work = rate × Time

therefore
30 = Rate × 6
or Rate = 30/6 = 5

Now we need to find the number of computers
total work given = 80 and total time = 3 hrs

therefore
80 = 5 × 3 × No. of computers
or No. of computers = 80/15 = 5.333 , but no. of computers cannot be fraction i.e , so we have to consider as 1.

Total no. of computers = 5 + 1 = 6.

Speed of the boat in still water = 8 km/hr
Speed upstream = 1/1 = 1 km/hr.
Speed of the stream = 8 - 1 = 7 km/hr.
Speed downstream = (8 + 7) = 15 km/hr.
Time is taken to travel 1 km downstream = (1/15) hr
= (1 × 60)/15
= 4 minutes.

Question: A train that is 300 meters long takes 30 seconds to pass a stationary pole. At the same constant speed, how much time will the train take to completely pass a platform that is 500 meters long? 

Solution:
Length of the train = 300 m
Time taken to pass a pole = 30 seconds

So, speed of the train
= 300/30
= 10 m/s

Length to be covered while passing the platform
= Length of train + Length of platform
= 300 + 500
= 800 m

Time taken to pass the platform
= Total distance/Speed
= 800/10
= 80 seconds

Therefore, the train will take 80 seconds to pass the platform.

Question: A train 150 meters long passes a signal post in 15 seconds. How long will it take to pass a bridge that is 450 meters long?

Solution:
Train's speed = Distance/Time
= 150/15 = 10 m/s

Total distance to pass the bridge,
= Length of train + Length of bridge
= 150 m + 450 m
= 600 m

∴ Required time = Distance/Speed
= 600/10
= 60 seconds
​= 1 minute

∴ The train will take 60 seconds or 1 minute to pass platform.

Question: 45 toymakers can prepare 30 toys per day. Rifat wants 360 toys. How many toymakers should he employ to get the job done in 12 days?

Solution:
Let, the required number of toymakers x
45 toymakers make 30 toys per day
So, 1 toymaker makes = 30/45 = 2/3 toys per day
Each toymaker in 12 days makes = (2/3) × 12 = 8 toys
So, x toymakers will make = 8x toys

ATQ,
8x = 360
⇒ x = 360 × (1/8)
∴ x = 45

Question: Speed ​​of the boat and current are 12 and 4 km/h respectively. How much time will it take for the boat to travel 64 km downstream and then return the same distance upstream?

Solution:
We know,
Effective speed with the current = Actual speed + speed of stream
= (12 + 4)  km/h
= 16 km/h

∴ Time taken to cover 64 km = (64 ÷ 16) hours 
= 4 hours

Effective speed against the current = Actual speed - speed of stream
= (12 - 4) km/h
= 8 km/h

∴  Time taken to return 64 km = (64 ÷ 8) hours
= 8 hours

∴ Total time taken = (4 + 8) hours 
= 12 hours

Previous speed = x
A/Q,
32/x - 32/(x + 4) = 4
Or, 32x + 128 - 32x/x(x + 4) = 4
Or, (x + 4)x = 32
Or, x² + 4x - 32 = 0
Or, x² + 8x - 4x - 32 = 0
Or, (x + 8) (x - 4) = 0
So, the previous speed was 4 kmph.
And, Present speed will be 4 + 4 = 8 kmph.

Let the length of the tunnel = x meter
Then, distance = (800 + x) meter.
Time = 1 minute = 60 seconds.
Speed = 78 km/hr
= 78 × (5/18)
= (65/3) m/s
According to the question,
800 + x = 60 × (65/3)
⇒ 800 + x = 1300
⇒ x = 500 meter.

Let x km/hr be the speed of the train.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)((360/x)- 1) = 360
⇒ (x + 5)(360 – x) = 360x
⇒ 360x – x² + 1800 - 5x = 360x
⇒ x² + 5x – 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ x = 40, -45

Negative value is not considered for speed, hence the answer is 40km/hr.

Downstream speed = 55/(5/2) = 11 × 2
= 22 km/hours

Time taken in upstream = 2.2 × 5/2
= 5.5 hours

Upstream speed = 55/5.5
= 10 km/hour

∴ The speed of boat in still water

= (10 + 22)/2
= 32/2
= 16 km/hr.

যেহেতু ট্রাক দুটি বিপরীত দিকে চলছে, সেহেতু এদের গতিবেগ যোগ করলে আপেক্ষিক বেগ পাওয়া যাবে। 

∴ নির্ণেয় সময় = 300 / (70+50) ঘণ্টা
= 300/120 ঘণ্টা
= 2.5 ঘণ্টা

Question: An aeroplane covers a certain distance at a speed of 250 kmph in 4 hours. To cover the same distance in hours, it must travel at a speed of:

Solution:
বিমানটির অতিক্রান্ত  মোট দূরত্ব = গতিবেগ × সময়
= 250 কিমি/ঘন্টা × 4 ঘন্টা
= 1000 কিমি

এখন, একই দূরত্ব ঘন্টায় অতিক্রম করার জন্য প্রয়োজনীয় গতিবেগ নির্ণয় করতে হবে।

ঘন্টা = (1 + 1/4) ঘন্টা = 5/4 ঘন্টা

প্রয়োজনীয় গতিবেগ = মোট দূরত্ব/সময়
= 1000 কিমি/(5/4) ঘন্টা
= (1000 × 4/5) কিমি/ঘন্টা
= 200 × 4 কিমি/ঘন্টা
= 800 কিমি/ঘন্টা

সুতরাং, একই দূরত্ব 5/4 ঘন্টায় অতিক্রম করার জন্য বিমানটিকে 800 কিমি/ঘন্টা গতিবেগে চলতে হবে।

Time taken by Javed =100/{18 × (5/18)} = 20 seconds
Time taken by Naveed = 20 + 5 = 25 seconds
Speed of Naveed = 100/25 × 18/5 = 14.4 km/h