Answer:
50 km/hr.
B
Explanation:
I get the same answer (50 km/hour) but I did it slightly differently and both solutions are worth seeing.
First of all you have to figure out how far the man runs. Assume he starts right at the tip of the cow catcher (the furthest point out on one of those old fashioned engines.
He runs at 5km / hour for 10 seconds.
5 km = 5000 meters.
5000 meters / hour * [1 hour / 3600 seconds ] = 1.38889 m/sec.
He does this for 10 seconds
d = r * t
d = 1.38889 * 10
d = 13.8889
Now look at what the train has to do. It passes him in 10 seconds. (The train has gone from the tip of the cow catcher to the end of the caboose in 10 seconds.)
d = 125 + 13.8889 meters
d = 138.8889 meters.
Now we have to convert this to km / hour
138.8889 m / 10 seconds [ 1 km/ 1000 m] * [ 3600 sec / 1 hr.]
(138.8889 * 1 * 3600 ) / (10 * 1000 * 1 )
50.000004
So the answer is 50 km/hr.