really hard quesiton

Question

A parked car is facing northward. A spider is walking southwest on the roof of the car at a speed of 1 meter per second when the car starts moving forward at a rate of 12 meters per second. What is the resulting speed and direction that the spider is moving relative to the ground?

Answer 1

You need to sum two velocity vectors.

 

Take north as up (positive y axis) and east as right (positive x axis)

 

The car's velocity vector, in coordinate form, is (0,12).

 

The spider's velocity relative to the car is 1 southwest, which is proportional to (-1,-1).  However,

(-1,-1) has a magnitude of sqrt(2), so we must divide by that to get the spider's coordinate velocity.

 

The spider's velocity vector relative to the car is (-1/sqrt(2),-1/sqrt(2))=(-0.707,-0.707)

 

The spider's velocity relative to the ground, v, is the sum of these velocity coordinates.

 

v=(-0.707,12-0.707)=(-0.707,11.293)

 

Speed is the magnitude of velocity, so speed, s, is

 

s²=(-0.707)²+(11.293)²=(1/2)+(11.293)²=128.03

 

s=11.315

 

The spider's speed is 11.315

 

The velocity vector (-0.707,11.293) lies in quadrant II.  This corresponds to a reference angle, B such that

 

tan(B)=11.293/0.707 =16

 

B=86.4 degrees

 

So, the spider's direction relative to the ground is 90-86.4=3.6 degrees west of north.

 

Using the bearing system, this is a track of 360-3.6=356.4 degrees.

 

Do the computations yourself to make sure there are no numerical errors.