A locomotive engine, without any wagons attached to it, can go at a speed of 40 km/hr. Its speed is diminished by a quantity that varies proportionally as the square root of the number of wagons attached. With 16 wagons, its speed is 28 km/hr. The
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Let x is the factor by which speed is diminished.
A/C to question, x is directly proportional to square root of wagons.
Let number of wagons is w.
then, , K is proportionality constant.
from the question, speed is decreased by 12(x) with 16 wagons.
so, 12 = K(√16) = 4K
K = 3
as speed reduced by 40 km/h then engine cannot move.
so, 40 = 3√w
w = (40/3)² = 1600/9 = 177.77 ≈ 178
hence, there are 178 weapons , engine cannot move. hence, maximum wagons = 177
A/C to question, x is directly proportional to square root of wagons.
Let number of wagons is w.
then, , K is proportionality constant.
from the question, speed is decreased by 12(x) with 16 wagons.
so, 12 = K(√16) = 4K
K = 3
as speed reduced by 40 km/h then engine cannot move.
so, 40 = 3√w
w = (40/3)² = 1600/9 = 177.77 ≈ 178
hence, there are 178 weapons , engine cannot move. hence, maximum wagons = 177
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# Complete question will be-
A locomotive engine without a train can run at 40 kmph. The speed diminishes by a quantity which varies as the square root of the number of wagons attached. With 16 wagons the speed is 28 kmph. The greatest number of wagons that the engine can move is...
# Answer - 177
# Explanation-
speed diminished=k × √wagons attached
For given condition,
40-28=k√16
k=3
For maximum no of wagons attached speed diminished will be close to but less than 40.
Now,
diminished speed<40
3×√wagons attached<40
√wagons attached<40/3
wagons attached < 1600/9=177.77
Rounding off to lower integer we get 177.
Maximum 177 wagons can be attached to the train.
A locomotive engine without a train can run at 40 kmph. The speed diminishes by a quantity which varies as the square root of the number of wagons attached. With 16 wagons the speed is 28 kmph. The greatest number of wagons that the engine can move is...
# Answer - 177
# Explanation-
speed diminished=k × √wagons attached
For given condition,
40-28=k√16
k=3
For maximum no of wagons attached speed diminished will be close to but less than 40.
Now,
diminished speed<40
3×√wagons attached<40
√wagons attached<40/3
wagons attached < 1600/9=177.77
Rounding off to lower integer we get 177.
Maximum 177 wagons can be attached to the train.
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