A train has length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.

Updated: 9 months ago
āωāĻ¤ā§āϤāϰāσ
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āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋāĻŦ⧇āĻ— (Relative Velocity)

āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋāĻŦ⧇āĻ— āĻšāϞ⧋ āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁āϰ āĻŦ⧇āĻ— āĻ…āĻ¨ā§āϝ āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁āϰ āϤ⧁āϞāύāĻžā§Ÿ āĻ•āϤ āϤāĻž āύāĻŋāĻ°ā§āĻĻ⧇āĻļ āĻ•āϰ⧇āĨ¤ āĻ…āĻ°ā§āĻĨāĻžā§Ž, āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁ āĻĨ⧇āϕ⧇ āĻ…āĻ¨ā§āϝ āĻŦāĻ¸ā§āϤ⧁āϕ⧇ āĻĒāĻ°ā§āϝāĻŦ⧇āĻ•ā§āώāĻŖ āĻ•āϰāϞ⧇ āϝ⧇ āĻŦ⧇āĻ— āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ, āϏ⧇āϟāĻŋāχ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋāĻŦ⧇āĻ—āĨ¤

āĻŽā§‚āϞ āϧāĻžāϰāĻŖāĻž

āϧāϰāĻž āϝāĻžāĻ• āĻĻ⧁āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁ A āĻāĻŦāĻ‚ B āϚāϞāϛ⧇āĨ¤ A-āĻāϰ āϤ⧁āϞāύāĻžā§Ÿ B-āĻāϰ āĻŦ⧇āĻ—āϕ⧇ āĻŦāϞāĻž āĻšā§Ÿ B-āĻāϰ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— (relative velocity of B with respect to A)āĨ¤

āϏ⧂āĻ¤ā§āϰ

āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— āύāĻŋāĻ°ā§āĻ­āϰ āĻ•āϰ⧇ āĻĻāĻŋāϕ⧇āϰ āωāĻĒāϰāĨ¤

ā§§. āĻāĻ•āχ āĻĻāĻŋāϕ⧇ āϚāϞāϞ⧇ (Same Direction)

V = V 1 - V 2

āϝ⧇āĻ–āĻžāύ⧇ V₁ > V₂ āĻšāϞ⧇ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻšā§ŸāĨ¤

āωāĻĻāĻžāĻšāϰāĻŖ:

āĻāĻ•āϟāĻŋ āĻ—āĻžā§œāĻŋ 60 km/h āĻāĻŦāĻ‚ āφāϰ⧇āĻ•āϟāĻŋ 40 km/h āĻŦ⧇āϗ⧇ āĻāĻ•āχ āĻĻāĻŋāϕ⧇ āϚāϞāϛ⧇āĨ¤
āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— = 60 − 40 = 20 km/h

⧍. āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻāĻŋāϕ⧇ āϚāϞāϞ⧇ (Opposite Direction)

V = V 1 + V 2

āωāĻĻāĻžāĻšāϰāĻŖ:

āĻĻ⧁āϟāĻŋ āĻŸā§āϰ⧇āύ 50 km/h āĻāĻŦāĻ‚ 30 km/h āĻŦ⧇āϗ⧇ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻāĻŋāϕ⧇ āϚāϞāϛ⧇āĨ¤
āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— = 50 + 30 = 80 km/h

ā§Š. āĻ¸ā§āĻĨāĻŋāϰ āĻĒāĻ°ā§āϝāĻŦ⧇āĻ•ā§āώāϕ⧇āϰ āĻ•ā§āώ⧇āĻ¤ā§āϰ⧇

āϝāĻĻāĻŋ āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁ āĻ¸ā§āĻĨāĻŋāϰ āĻĨāĻžāϕ⧇, āϤāĻŦ⧇ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— = āϚāϞāĻŽāĻžāύ āĻŦāĻ¸ā§āϤ⧁āϰ āĻŦ⧇āĻ—āĨ¤

āϗ⧁āϰ⧁āĻ¤ā§āĻŦāĻĒā§‚āĻ°ā§āĻŖ āϤāĻĨā§āϝ

â€ĸ āĻāĻ•āχ āĻĻāĻŋāϕ⧇ → āĻŦāĻŋā§Ÿā§‹āĻ— āĻšā§Ÿ
â€ĸ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻāĻŋāϕ⧇ → āϝ⧋āĻ— āĻšā§Ÿ
â€ĸ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻŦ⧇āĻ— āĻļ⧁āϧ⧁ āĻĻāĻŋāϕ⧇āϰ āωāĻĒāϰ āύāĻŋāĻ°ā§āĻ­āϰ āĻ•āϰ⧇

āĻŽāύ⧇ āϰāĻžāĻ–āĻžāϰ āĻ•ā§ŒāĻļāϞ

â€ĸ Same direction → subtract
â€ĸ Opposite direction → add

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āωāĻ¤ā§āϤāϰāσ

In 1 hour, one train goes (21-16)= 5 km more than another.

Now, In 1 hour total distance covered by them = (16+21) km = 37 km

If distance difference is 5 km then total distance = 37 km

If distance difference is 60 km then total distance = 37×605=444 km 

Distance between Dhaka and Chittagong is 444 km. (ans.)

Tamanna
Tamanna
2 years ago
1.3k
āωāĻ¤ā§āϤāϰāσ

Here, relative speed of Y in terms of X

=4x3-x=4x-3x3=x3 mph.

X will cover in (11.30 - 10) = 1.5 hours = ( x Ã— 1.5) miles = 1.5x miles

∴Required time taken by Y to catch X= 1.5xx3=1.5x×3x=4.5x hours

∴Y will meet X at =11.30 am +4.5 hour = 4 pm

Alternative: Given that,

X' s speed is = x mph

∴Y's speed is =4x3 mph

Let, X and Y will meet after p hours

According to questions, px = (p - 1.5) Ã— 4x3 [As, d = time Ã— velocity]

=px=4px3-2x =4px3-px=2x =4px-3px3=2x =px=6x âˆ´p=6 âˆ´Y will catch train X at = 10 am +6 hours = 4pm (answer)

PRONAY TIRKI
PRONAY TIRKI
2 years ago
1k
āĻļāĻŋāĻ•ā§āώāĻ•āĻĻ⧇āϰ āϜāĻ¨ā§āϝ āĻŦāĻŋāĻļ⧇āώāĻ­āĻžāĻŦ⧇ āϤ⧈āϰāĻŋ

ā§§ āĻ•ā§āϞāĻŋāϕ⧇ āĻĒā§āϰāĻļā§āύ, āĻļā§€āϟ, āϏāĻžāĻœā§‡āĻļāύ āĻ“
āĻ…āύāϞāĻžāχāύ āĻĒāϰ⧀āĻ•ā§āώāĻž āϤ⧈āϰāĻŋāϰ āϏāĻĢāϟāĻ“āϝāĻŧā§āϝāĻžāϰ!

āĻļ⧁āϧ⧁ āĻĒā§āϰāĻļā§āύ āϏāĻŋāϞ⧇āĻ•ā§āϟ āĻ•āϰ⧁āύ — āĻĒā§āϰāĻļā§āύāĻĒāĻ¤ā§āϰ āĻ…āĻŸā§‹āĻŽā§‡āϟāĻŋāĻ• āϤ⧈āϰāĻŋ!

āĻĒā§āϰāĻļā§āύ āĻāĻĄāĻŋāϟ āĻ•āϰāĻž āϝāĻžāĻŦ⧇
āϜāϞāĻ›āĻžāĻĒ āĻĻ⧇āϝāĻŧāĻž āϝāĻžāĻŦ⧇
āĻ āĻŋāĻ•āĻžāύāĻž āϝ⧁āĻ•ā§āϤ āĻ•āϰāĻž āϝāĻžāĻŦ⧇
Logo, Motto āϝ⧁āĻ•ā§āϤ āĻšāĻŦ⧇
āĻ…āĻŸā§‹ āĻĒā§āϰāϤāĻŋāĻˇā§āĻ āĻžāύ⧇āϰ āύāĻžāĻŽ
āĻ…āĻŸā§‹ āϏāĻŽāϝāĻŧ, āĻĒā§‚āĻ°ā§āĻŖāĻŽāĻžāύ
āĻĒā§āϰāĻļā§āύ āĻāĻĄāĻŋāϟ āĻ•āϰāĻž āϝāĻžāĻŦ⧇
āϜāϞāĻ›āĻžāĻĒ āĻĻ⧇āϝāĻŧāĻž āϝāĻžāĻŦ⧇
āĻ āĻŋāĻ•āĻžāύāĻž āϝ⧁āĻ•ā§āϤ āĻ•āϰāĻž āϝāĻžāĻŦ⧇
Logo, Motto āϝ⧁āĻ•ā§āϤ āĻšāĻŦ⧇
āĻ…āĻŸā§‹ āĻĒā§āϰāϤāĻŋāĻˇā§āĻ āĻžāύ⧇āϰ āύāĻžāĻŽ
āĻ…āĻŸā§‹ āϏāĻŽāϝāĻŧ, āĻĒā§‚āĻ°ā§āĻŖāĻŽāĻžāύ
āĻ…āĻŸā§‹ āύāĻŋāĻ°ā§āĻĻ⧇āĻļāύāĻž (āĻāĻĄāĻŋāϟāϝ⧋āĻ—ā§āϝ)
āĻ…āĻŸā§‹ āĻŦāĻŋāώāϝāĻŧ āĻ“ āĻ…āĻ§ā§āϝāĻžāϝāĻŧ
OMR āϏāĻ‚āϝ⧁āĻ•ā§āϤ āĻ•āϰāĻž āϝāĻžāĻŦ⧇
āĻĢāĻ¨ā§āϟ, āĻ•āϞāĻžāĻŽ, āĻĄāĻŋāĻ­āĻžāχāĻĄāĻžāϰ
āĻĒā§āϰāĻļā§āύ/āĻ…āĻĒāĻļāύ āĻ¸ā§āϟāĻžāχāϞ āĻĒāϰāĻŋāĻŦāĻ°ā§āϤāύ
āϏ⧇āϟ āϕ⧋āĻĄ, āĻŦāĻŋāώāϝāĻŧ āϕ⧋āĻĄ
āĻ…āĻŸā§‹ āύāĻŋāĻ°ā§āĻĻ⧇āĻļāύāĻž (āĻāĻĄāĻŋāϟāϝ⧋āĻ—ā§āϝ)
āĻ…āĻŸā§‹ āĻŦāĻŋāώāϝāĻŧ āĻ“ āĻ…āĻ§ā§āϝāĻžāϝāĻŧ
OMR āϏāĻ‚āϝ⧁āĻ•ā§āϤ āĻ•āϰāĻž āϝāĻžāĻŦ⧇
āĻĢāĻ¨ā§āϟ, āĻ•āϞāĻžāĻŽ, āĻĄāĻŋāĻ­āĻžāχāĻĄāĻžāϰ
āĻĒā§āϰāĻļā§āύ/āĻ…āĻĒāĻļāύ āĻ¸ā§āϟāĻžāχāϞ āĻĒāϰāĻŋāĻŦāĻ°ā§āϤāύ
āϏ⧇āϟ āϕ⧋āĻĄ, āĻŦāĻŋāώāϝāĻŧ āϕ⧋āĻĄ
āĻāĻ–āύāχ āĻļ⧁āϰ⧁ āĻ•āϰ⧁āύ āĻĄā§‡āĻŽā§‹ āĻĻ⧇āϖ⧁āύ
ā§Ģā§Ļ,ā§Ļā§Ļā§Ļ+
āĻļāĻŋāĻ•ā§āώāĻ•
ā§Šā§Ļ āϞāĻ•ā§āώ+
āĻĒā§āϰāĻļā§āύāĻĒāĻ¤ā§āϰ
āĻŽāĻžāĻ¤ā§āϰ ā§§ā§Ģ āĻĒ⧟āϏāĻžā§Ÿ āĻĒā§āϰāĻļā§āύāĻĒāĻ¤ā§āϰ
ā§§ āĻ•ā§āϞāĻŋāϕ⧇ āĻĒā§āϰāĻļā§āύ, āĻļā§€āϟ, āϏāĻžāĻœā§‡āĻļāύ āϤ⧈āϰāĻŋ āĻ•āϰ⧁āύ āφāϜāχ

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