How many tons of asphalt concrete will be required to overlay a 2.5 mile long 24 ft wide road with 3 in of wearing course asphalt concrete that has a density of 1479 pcf?
To determine the weight of the asphalt needed, first determine the area of the surface to be paved.
Since the shape of the road is a rectangle, use the formula of area of rectangle.
A `=` Length `xx` Width
Before plugging-in the values, convert the 2.5 miles to feet in order that the dimension to have same unit.
`2.5 mi xx (5280 ft)/(1mi) = 13200ft`
Now that the length is in feet, plug-in the values of dimension of the road to the formula of area of rectangle.
Hence, the area to be paved is 316 800 `ft^2` .
Next, determine the volume. Use the formula:
V `=` Area `xx` thickness
Before substituting the values, convert the thickness to feet in order to have same units.
`3 i n xx (1ft)/(12 i n)=0.25ft`
Then, substitute the values of the area and thickness to the formula of volume.
Thus, the volume of the asphalt needed is 79200 ft^3.
Next, to determine its weight , multiply that with the given density 1479 pounds per cubic feet (1479 lb/ft^3).
So, the weight is 11 7136 800 lbs.
Then, convert it to tons.
`117136800 lbs xx (1t o n)/(2000lbs) = 58568.4 t o n s`
Therefore, the amount of asphalt needed is 58 568.4 tons.
Steps: Find volume -> Find Pounds -> Find Tons
Convert everything to feet (3in=.25ft):
`(2.5 mi) * (5280 ft)/(1 mi) = 13200 ft`
`13200ft * .25ft * 24ft = 79200 ft^3`
Use density to convert from volume to pounds:
`(79200 ft^3) * (1479 lbs)/(1 ft^3)`
`= 117136800 lbs`
Convert from pounds to tons:
`117136800 lbs * (1 tn)/(2000 lbs) = 58568.4 tns`