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

a) The magnitud is 0.78m.

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

Step-by-step 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)})=√((-0.32)^2+(-0.72)^2) =0.78m`.

b) If
`\theta` is the direction of
`B_{2_(2)}[tex], we have:</p><p>[tex]tan(\theta )=(-0.72)/(-0.32)` ⇒
`\theta=arctan((-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°.