Discussion :: Problems on Trains - General Questions (Q.No.12)
m/sec
x + 250 = 520| Nisha Dahiya said: (Sep 21, 2010) | |
| Why 72 is multiplied by 5/18? | |
| Sania said: (Oct 4, 2010) | |
| @ NISHA Speed is given 72 kmps. But length of platform is in meters. so we have to change it into m/s. |
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| Sonam said: (Nov 11, 2010) | |
| Ya. But why only 5/18 why not some other numbers? | |
| Dinesh Guptha said: (Nov 16, 2010) | |
| Hi sonam, Let me explain about converting km/hr into m/s in detail. 1 km = 1000 meters. 1 hr = 3600 seconds. 1 km/hr = 1000/3600 m/s = 10/36 m/s = 5/18 m/s. Therefore 1 km/hr = 5/18 m/sec. |
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| Donald Pinto said: (Dec 3, 2010) | |
| Hi Dinesh Thankyou so much for the explanation!! | |
| Ashok.K.Rathod said: (Dec 20, 2010) | |
| Thanks dinesh. | |
| Sonia said: (Jan 23, 2011) | |
| why x+250?.....for same direction we take sbtraction....then why it is added? | |
| Yogesh said: (Mar 13, 2011) | |
| x + 250 = 520, how 520 has come, please explain. | |
| Savi said: (Mar 18, 2011) | |
| Hi @Yogesh the formula is length=speed*time here l=x*250 speed=72*5/18=20ms time=26s Apply formula x+250=20*26 =520 Now got it |
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| Akber said: (Mar 25, 2011) | |
| @Sania. Here platform is stationary that is the reason we are adding. |
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| Raj said: (Dec 26, 2011) | |
| @Savi here l=x*250 or l=x+250 ? |
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| Sujith.M said: (Jan 15, 2012) | |
| (x+250)/26=20 Can anybody explain how to come this? | |
| Kunal Kanchan Srivastava said: (Apr 8, 2012) | |
| Station is crossed when tail of train pass, so we can't take length of train be x it should be 2x and length of the train must be 135m. | |
| Naresh said: (Jun 17, 2012) | |
| Here they given train speed=72kmph now convert into m/s that is 72*5/18=20m/s. And time 26sec given. This time taken by the train to cross the total platform. So now we know train speed, and time. Speed=distance/time. Distance=speed*time=20*26=520m. So this is total distance of platform and train. i.e (platform+train length). Platform+Train length=520. 250+Train length=520. Train length=520-250=270m. |
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| Manovasanth said: (Jul 23, 2012) | |
| @Sonia. Here relative speed in sense where two forces come in opposite directions and in turn of collide force. In above said question the trans come in same direction so it is added if it in opposite direction it is subtracted. |
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| Benhur Dsilva said: (Oct 23, 2012) | |
| First convert 72km/hr in m/sec So, 72/1*1000/3600 =72*5/18 =20m/sec. D = s * t = 20*26 = 520. Its a distance of Train + platform; this is because full length of train pass full length of platform. So, 520-250=270. |
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| Saurabh said: (Jan 18, 2013) | |
| Speed in meters = 72 * (5/18) = 20. Length = speed * time. 250 + length of platform = 20*26. Length of platform = 520-250 = 270. |
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| Sandip said: (Mar 22, 2013) | |
| At 72 km/hr train goes 520 mt. Length of the plat form = 250. So length of the train = 520-250 = 270. |
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| Rakesh said: (Feb 4, 2014) | |
| Distance = speed * Time. So Distance = X+250. Where x= Length of train. Speed = 72*5/18 m/s (Conversion from km/s to m/s). i.e., 20 m/s. And time = 26 seconds. Therefore X+250 = 20 * 26. X = 520-250. X = 270 m. |
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| Suman said: (Feb 18, 2014) | |
| Train speed m/s = 72x5/18 = 20 sec/m. Length of train = speed x time. = 26*20 = 520m - 250(length of the platform). So train length = 270m. Is it right or wrong? |
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| Jeevan said: (May 12, 2014) | |
| = 72*5/18. = 36*5/9. = 4*5. = 20. d = Sp*ti. 20*26. = 520-250. = 270. |
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| Judge said: (Jun 11, 2014) | |
| What is SP and TI? | |
| Mudassir Pasha said: (Jul 8, 2014) | |
| Shortcut: Speed of the train is 20 m/sec. Therefore for 1 sec--20 m. For 26 sec -- ? 20*26=520/1. It means for 26 sec train crosses 520m. It includes length of train also, therefore we need minus the length of the train to get the length of the platform. i.e., 520-250 = 270. |
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| Sakshi said: (Feb 4, 2017) | |
| I still didn't understand why we are adding the length of the train with the distance travelled? Please anyone explain it. |
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| Sanket Gawai said: (Feb 25, 2017) | |
| I'll explain you @Sakshi that why we are adding with the distance travelled. Suppose the train passes through a pole or a stationary man we consider only length of the train as a distance travelled because length of man or pole is negligible but when we come across the platform well have to consider the length of platform also as a distance travelled because the train passes the platform in the given time so lengthy of platform is also added with the length of train. |
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| Piyush said: (Mar 24, 2017) | |
| The formula used is : time taken by the train = (length of train + length of platform)/ speed of the train. Where the length of the train is taken as X. |
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| Gaurav Tanwar said: (May 31, 2017) | |
| Why x+250 is taken? Explain. | |
| Jyoti said: (Jun 14, 2017) | |
| Thank you so much for all your explanations. | |
| Sarang said: (May 4, 2018) | |
| Distance =speed *time. D=72km/h*26, D=72*5/18*26, D=4*5*26, D=20*26, D=520-250=270m. |
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| Jameel said: (Jun 25, 2018) | |
| Thanks for your solution @Sarang. | |
| Ronik Patel said: (Oct 17, 2018) | |
| @All. Why 72 is multiplied by 5/18? Ans: Because of unit conversion..... 72 is in kmph and our answer is in the meter so it is required to convert speed from kmph to mph. I hope, this will help you all. |
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| Ashok said: (Dec 1, 2018) | |
| Use this Formula for all train problems. Train length + object length * hr * 60m * 60se/speed(1000). 26= T+250 * 60 * 60/72000, 250+ 26 * 72000/60 * 60, 520 - 250 = 270. |
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