Assume that the distance that a car runs on one liter of petrol varies inversely as the square of the speed at which it is driven. It give a run of 25 kilometer per liter at speed of 30 kilometer per liter. At what speed should it be driven to get a run of 36 kilometer per liter.
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mileage of the car = m km / litre
speed of the car = v km/hr
distance driven = s km and amount of petrol = L litres
mileage is inverse proportional to the square of the speed.
m α 1/v²
m = K / v², where K = constant of proportionality , units: km/litre * km²/hr²
we know, 25 km/litre = K / (30 kmph)²
K = 25 * 900 = 22,500 km³/litre-hr²
m = 22, 500 /v²
Now, v² = K / m = 22, 500 / 36
v = 150/6 = 25 km/hr
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without finding the constant of proportionality:
m2 / m1 = v1² / v2²
m1 = 25 km/litre v1 = 30 kmph
v2 = ? m2 = 36 km/litre
v2² = m1 v1² / m2 = 25 * 30² / 36 =25 * 25
v2 = 25 kmph
speed of the car = v km/hr
distance driven = s km and amount of petrol = L litres
mileage is inverse proportional to the square of the speed.
m α 1/v²
m = K / v², where K = constant of proportionality , units: km/litre * km²/hr²
we know, 25 km/litre = K / (30 kmph)²
K = 25 * 900 = 22,500 km³/litre-hr²
m = 22, 500 /v²
Now, v² = K / m = 22, 500 / 36
v = 150/6 = 25 km/hr
============================
without finding the constant of proportionality:
m2 / m1 = v1² / v2²
m1 = 25 km/litre v1 = 30 kmph
v2 = ? m2 = 36 km/litre
v2² = m1 v1² / m2 = 25 * 30² / 36 =25 * 25
v2 = 25 kmph
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