1. A-20 km south east
The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:
[tex]d_x = 12 km[/tex] east
[tex]d_y = 16 km[/tex] south
Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km[/tex]
and the direction is between the two original directions, so south-east.
2. D. 10 m/s
First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have
[tex]S=\frac{1}{2}gt^2[/tex]
From which we find the total time of the fall, t:
[tex]t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s[/tex]
Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is
[tex]v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s[/tex]
3. B=32.32 m
As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have
[tex]t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s[/tex]
And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is
[tex]d=vt=(16 m/s)(2 s)=32 m[/tex]
So, the closest answer is B.
