Question

Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.41 m due east, then 0.72 m at 38o north of due east. Beetle 2 also makes two runs; the first is 1.74 m at 48o east of due north. What must be (a) the magnitude and (b) the direction (relative to the east direction in the range of (-180°, 180°)) of its second run if it is to end up at the new location of beetle 1?

Answer 1

Answer:

a) The magnitud is 0.78m.

b) The direction is $-113.97^\circ$.

Explanation:

Beetle 1 runs 0.41m due east, then 0.72m at $38^\circ^$ north of due east.

This means that it's vector position at the end for Beetle 1 is (in vector notation):

$B_1=(0.41m ; 0) + (0.72m.cos(38^\circ ) ; 0.72m.sin(38^\circ )) =  (0.41m+0.72m.cos(38^\circ ) ; 0.72m.sin(38^\circ ))$

⇒ $B_1=(0.97m ; 0.44m) =(0.97 ; 0.44)m$

Beetle 2 first run is 1.74m at $48^\circ$ east of due north, this is:

$B_{2_{1}}=(1.74m.sin(48^\circ ) ; 1.74m.cos(48^\circ ) ) = (1.29 ; 1.16)m$

What's the difference between the final position of Beetle 1 and the position of the Beetle 2 after the first run? This difference will be the vector related to the second run of Beetle 2. This will be:

$B_{2_{2}}=B_1 - B_{2_{1}}= (0.97 ; 0.44)m - (1.29 ; 1.16)m$

⇒ $B_{2_{2}}=(-0.32 ; -0.72)m$

a) The magnitud the this second run will be:

$M(B_{2_{2}})=\sqrt{(-0.32)^2+(-0.72)^2} =0.78m$.

b) If $\theta$ is the direction of $B_{2_{2}}[tex], we have:

[tex]tan(\theta )=\frac{-0.72}{-0.32}$ ⇒ $\theta=arctan(\frac{-0.72}{-0.32})$ ⇒ $\theta = 66.03^\circ$.

But from $B_{2_{2}}$ components we know the vector points out to the third quadrant (the angle gave of that value because signs canceled each other). The actual direction is

Ф=$\theta$ - 180° = -113.97°.