We use the formula, to calculate the volume of water displaced by concrete canoe,
[tex]V =\frac{W}{\gamma }[/tex]
Here, W is the weight of concrete canoe and [tex]\gamma[/tex] is the specific weight of water and its value is [tex]62.4\ lb/ft^3[/tex].
So,
[tex]V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3[/tex].
Now the volume of water occupied in ultra lightweight kevlar canoe,
[tex]v=\frac{w}{\gamma}[/tex]
Here, w is weight of kevlar canoe.
So,
[tex]v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3[/tex]
Thus, the volume of water displaced,
[tex]=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3[/tex].
Hence, the volume of water displaced canoe compared to an ultra-lightweight kevlar canoe is [tex]2.19\ ft^3[/tex]