# 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?

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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.

So,

`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.

`A=13200*24`

`A=316800`

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.

So,

`3 i n xx (1ft)/(12 i n)=0.25ft`

Then, substitute the values of the area and thickness to the formula of volume.

`V=316800*0.25=79200`

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).

`W=79200*1479=117136800`

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`

Find Volume:

`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`