on a 100 km track ,a train travels the first 30 kmat uniform speed of 30 km/ h. how fast must the train travel next 70 km, so that the average speed of train for the entire trip becomes 40 km/h?
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Given: d1 = 30
d2 = 70
we know that s = d/t and avg speed = total distance / total time taken
when the train acquired a speed of 30km/h, the time taken is -
30 = 30/t .... 30/30 = t1 = 1 second
Now we know that the avg speed = total distance upon total time taken
= d1 + d2 / t1 + t2
[ where t2 = time for the second part, for 70 km ]
40 = 100 / 1 +t2
40 + 40t2 = 100
40t2 = 100 - 40 = 60
t2 = 1.5 hours
Speed for the second part is = d2/t2
s = 70
. . . 1.5
s = 46.666.. km/h or 46.7km/h
hope this help u buddy.....
d2 = 70
we know that s = d/t and avg speed = total distance / total time taken
when the train acquired a speed of 30km/h, the time taken is -
30 = 30/t .... 30/30 = t1 = 1 second
Now we know that the avg speed = total distance upon total time taken
= d1 + d2 / t1 + t2
[ where t2 = time for the second part, for 70 km ]
40 = 100 / 1 +t2
40 + 40t2 = 100
40t2 = 100 - 40 = 60
t2 = 1.5 hours
Speed for the second part is = d2/t2
s = 70
. . . 1.5
s = 46.666.. km/h or 46.7km/h
hope this help u buddy.....
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