At an antique car rally, a Stanley Steamer automobile travels north at 38 km/h and a Pierce Arrow automobile travels east at 51 km/h. Relative to an observer riding in the Stanley Steamer, what are the x- and y-components of the velocity of the Pierce Arrow car?
V of pa relative to ss
+ V of ss relative to ground
= V of pa relative to ground
Therefore (using an abbreviated notation)
Vpa/ss = Vpa/g - Vss/g
= 51 i - 38 j
i is a unit vector east and j is a unit vector north.
The x component is 51 km/h and the y component is -38
To find the x- and y-components of the velocity of the Pierce Arrow relative to the Stanley Steamer, we can use vector addition by breaking down the velocities into their respective components.
Let's assume the positive x-direction is east and the positive y-direction is north, as given in the question.
The velocity of the Stanley Steamer is given to be 38 km/h in the north direction. Since it is traveling straight north, its velocity in the x-direction is 0 km/h.
The velocity of the Pierce Arrow is given to be 51 km/h in the east direction. Since it is traveling straight east, its velocity in the y-direction is 0 km/h.
So, the x-component of the velocity of the Pierce Arrow relative to the Stanley Steamer is 51 km/h (as it is in the positive x-direction) and the y-component of the velocity is -38 km/h (as it is in the negative y-direction).
Therefore, the x-component of the velocity of the Pierce Arrow relative to the Stanley Steamer is 51 km/h, and the y-component is -38 km/h.