Question

Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.41 m due east, then 0.81 m at 22^o north of due east. Beetle 2 also makes two runs; the first is 1.39 m at 45^o 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

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Diagram 1 is a rough vector representation of position coordinate of beetle 1 .

OA represents ist run of Beetle 1 in due east dirrction and AB is its second run .

Finding final coordinates of Beetle 1

X1 = OA + AB cos 22°

X1 = .41+ 0.81 cos 22°

X1= 1.161 m

Y1 = 0+ 0.81sin22°

Y1= 0.3034m

Diagram 2 is a vector representation of position of beetle 2

Now

X2 = 1.39 cos45°

X2= 0.982m

Y2 = 0.982m

Above is a Rough combined

Diagram vector diagram of 1 & 2

Now from combined diagram it is clear that Beetle 2 have to run along CB to reach the new final position of Beetle 1

So,

Distance in x-direction ( x3) remained to cover to reach at B

= X1 - X2

= 1.161 - 0.982

= 0.179 m..........eqn (1)

Distance (y3) in y direction need to cover to reach B for Beetle 2

= Y1 - Y2

= 0.3034 - 0.982

= (- 0.6786) m...........eqn (2)

Let vector CB makes makes $\dpi{150} \bg_black \Theta$ with east direction

( x direction)

tan$\dpi{150} \bg_black \Theta$ = y3/ x3 ( from eqn 1& 2)

tan$\dpi{150} \bg_black \Theta$ = -(0.6786/0.179)

$\dpi{150} \bg_black \Theta$ = 75.22° ( Clockwise direction)

Hence

Final vector of Beetle running is

r = 0.179 m i^ - 0.6786J^

Its magnitude = √ ( 0.179^2 + ( -0.6786)^2 ( square root .....)

​​​​​​= 0.701 m

Hence :

a) Magnitude = 0.701m

b) 75.22° Clockwise relative to east direction and is reprensted as CB