A railroad track and a road cross at right angles. An observer stands on the road 70 meters south of the crossing and watches an eastbound train traveling at 60 meters per second. At how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection?

Question

savvy · Answer

57.6m/s

anonymous · Answer

4 seconds after he passes, he is 240 meters east of the intersection.  His velocity is 60m/s towards the east.   The vector from you towards the train is 240 E - 70 N.   The unit vector towards the train is (240E - 70N)/sqrt(240^2+70^2) = 240E/250 - 70N/250.
The velocity projected onto the unit vector is 60 * 240/250 = 57.6 m/s

savvy · Answer

nice one CeW....I did it using derivatives.....

anonymous · Answer

Thank you so much guys! :)