A horse can pull an empty cart at the speed of 18km per hour and the reduction in its speed is directly proportional to the square root of the number of boxes it can carry of equal weight of 10kg each. If the speed of the cart is 12km per hour when 9 boxes are loaded in the cart, find the maximum weight that can be carried if the speed of the cart is to be maintained at least 10km per hour.
Answers
Answer:
The maximum weight that can be carried when speed is 10 km/h is 120 Kg.
Step-by-step explanation:
Let n be the number of boxes loaded in the cart.
When the cart is empty the horse speed is 18 km/h and when 9 boxes are loaded the speed is 12 km/h.
Thus we can say that 18 - 12 = 6, and when the speed reduction is equal to 6 when know that n is equal to 9.
Since the reduction in its speed is directly proportional to the square root of the number of boxes, we have that 12 - 6 = 6, and the number of boxes loaded in the cart when the speed is 6 km/h is equal to = 81.
Then we can calculate the middle value of speed which is 9, and then number of boxes is . Now we know values when the speed is 9 and 12, and if we keep doing the middle values of speed and boxes we will conclude that when the speed is 10 km/h the number of boxes is, approximately, equal to 12.
Since boxes weight is 10 kg each we conclude that the maximum weight that can be carried when speed is 10 km/h is 12 x 10 = 120 Kg.
Answer:
The Horse can carry 16 boxes or 160Kgs for maintaining 10Kmph.
Step-by-step explanation:
Normal cart speed = 18kmph.
Reduction in speed is directly proportional to square root of number of boxes.
Let us say Number of boxes = n.
Reduction in speed = r
=> r = k√n
With 9 boxes harse / cart is going at 12kmph.
r = 18 - 12 = 6kmph. (reduction in speed)
n = 9. (Number of boxes.)
=> 6 = k√9 (substituting in above equ.)
=> k = 2.
Now for cart to go at 10kmph.
r = 18 - 10 = 8. (reduction in speed)
8 = 2√n (Substituting values in above equation)
=> √n = 8/2 = 4.
=> n = 4*4 = 16