Physics Asked by AdderallAstra on April 5, 2021
There’s a problem that I’ve been working on in physics for a couple of days now, and I’m just wondering if my thinking is wrong here. Say you’re given the velocity of an object (with an x- and y-component) relative to another object’s velocity (with only an x-component). If I’m trying to determine the first object’s y-component, then do I need to take into account the second object’s velocity? I’m able to get a logical answer using only trigonometry with the first object, and it makes sense to me that the x-component of the second object wouldn’t affect the y-component of the first object, but it still feels off for some reason.
If you are trying to work out the relative velocity (as in the title of your question) then the relative velocity between the two objects is a simple vector addition (subtraction) ie.,
$hat v_r = hat v_1 - hat v_2 $
$= (v_{1x} hat i + v_{1y} hat j) - (v_{2x} hat i + v_{2y} hat j)$
$ = (v_{1x} - v_{2x} ) hat i + (v_{1y} - v_{2y}) hat j$
And each object’s velocity components exist independently if you consider each object separately.
Answered by joseph h on April 5, 2021
No the first object's velocity in $Y - direction$ is not affected by the second object's velocity in the $X - direction$.
To understand why this happens, you must know what you mean by relative velocity .
Suppose two bodies $A$ and $B$ are moving in the same direction with somes velocities $V_A$ and $V_B$ as seen from ground frame ($V_A$ > $V_B$) and both moves for some time $t$ . For $B$ , he was at rest (but not from ground frame) and $A$ was moving with a constant velocity.
Now if you ask from $B$ about the distance travelled by $A$ he would say it to be $(V_A - V_B)t$ since every observer carries its own set of coordinates.
But if $A$ was travelling in upward direction, then it's vertical distance as seen from $B$ and from ground frame will be the same since $B$ never tried to move in that vertical direction. But if $B$ also had some vertical velocity, then the distance travelled by $A$ as seen from B and from ground frame would have been different . And in that case , you must take into account the vertical velocity of $B$.
Hope it helps ☺️.
Answered by A Student 4ever on April 5, 2021
Picture yourself travelling along the straight road and seeing somebody on the pavement who is stationary relative to the pavement/kerb/road.
You measure the perpendicular distance between the person on the pavement and the edge of the kerb.
That distance does not change with time.
If you now repeat the "experiment" but the person on the pavement moving at the same velocity as you.
You see the person as stationary relative to you and the perpendicular distance of the person from the kerb again stays the same.
Repeat the experiment with the person also moving at right angles to the edge of the kerb.
Whether you are moving along the road or not you will always measure the same component of velocity perpendicular to the edge of the kerb, relative to the pavement, of the person on the pavement.
So as measured, relative to the pavement/kerb/road, by you travelling along a road the component of the velocity at right angles to the kerb of a person moving on the pavement does not change.
Answered by Farcher on April 5, 2021
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