Describe motion from a moving observer

One cart, two observers

Imagine watching a cart from the ground and from a platform that moves steadily along the same line. You and the platform observer can describe different positions and velocities without disagreeing about which cart you watched. A frame specifies how positions and times are assigned. Here both observers use the same clock, metre unit and positive direction; their relative velocity is constant.

The widget prescribes the same ground trajectory for every setting: the cart goes from 0 to 6 m in three seconds, then returns to 3 m in another three seconds. Changing the observer is not an intervention that pushes the cart. The ground curve therefore stays fixed while the observer-relative curve changes.

Subtract the observer’s position

Let x(t) be the cart’s ground position at time t, a the observer’s ground position at time zero, and u the observer’s constant ground velocity. Time is in seconds, a and x are in metres, and u is in m/s. At time t the observer’s origin is at a + ut. The cart’s coordinate relative to that observer is:

x′(t) = x(t) − a − ut

The prime on x′ means “in the other frame,” not a derivative in this lesson. Read the equation as “subtract the observer’s current origin from the cart’s ground position.” The product ut has units of metres, so every term has the same unit. This is a one-dimensional Galilean coordinate transformation under the stated shared-clock and constant-velocity assumptions. See OpenStax relative motion for the broader frame context.

Start with a = 0 m and u = +0.5 m/s. At 3 s, the observer is at +1.5 m while the cart is at +6 m, giving relative position +4.5 m. At 6 s, both are at +3 m, giving relative position 0 m. The cart started at 0 m relative to this observer and ends there again.

Zero average velocity can coexist with travel

For this moving observer, the first segment velocity is 2−0.5 = +1.5 m/s and the second is −1−0.5 = −1.5 m/s. Relative displacement over six seconds is zero. Relative distance is 4.5+4.5 = 9 m. Average velocity is 0 m/s and average speed is 1.5 m/s.

Your action: capture this case and compare it with the ground-frame capture. Explain how zero endpoint change can coexist with nonzero distance, using the intermediate position. A statement such as “it did not move because its average velocity was zero” discards the route between the endpoints.

For these particular two frames, total distance happens to equal 9 m. Distance is not generally invariant between moving frames. Set u = −2 m/s: both relative segment velocities are positive, +4 and +1 m/s. The relative distance is now 12+3 = 15 m, equal to the positive displacement because there is no relative direction reversal.

Match one segment’s velocity

Set u = +2 m/s and a = 0 m. During the first segment the observer keeps pace with the cart, so relative velocity is 0 m/s and the dashed position curve is horizontal. During the second segment, relative velocity is −1−2 = −3 m/s. Over its three seconds that gives −9 m of relative displacement.

Now set u = −1 m/s. The second segment becomes stationary relative to the observer, while the first has relative velocity +3 m/s. A stationary-looking object can be moving relative to the ground. Conversely, an observer’s motion can make a ground-stationary object appear to move. Always state the reference frame before making a numerical velocity claim.

The general segment rule is v′ = v − u, where v is ground velocity and v′ is observer-relative velocity. Subtracting a constant origin position a does not affect slope. Subtracting the time-dependent term ut does. This is why moving an origin once and moving it continuously are different operations.

Bound the conclusion

The frames here share a positive axis; reversing that axis would introduce an additional sign change. They also share elapsed time, and the observer velocity is constant. Rotating frames, accelerating observers and speeds requiring relativistic treatment need additional ideas. The widget is not a model of how a real camera samples or distorts images.

The ground path has an ideal sharp corner at 3 s. Subtracting a smooth constant-velocity observer path does not remove its unequal left and right slopes. We do not invent one instantaneous velocity there. On either straight segment, however, the slope and the corresponding relative velocity are well defined.

Transfer: An object moves at +5 m/s relative to the ground for four seconds, starting at ground position +1 m. An observer starts at ground position +3 m and moves at +2 m/s. Find the object’s initial and final relative positions, relative displacement, distance and velocity. Keep the origin offset separate from the velocity subtraction.

Check the frame transfer

Initial relative position is 1−3 = −2 m. Final ground positions are 21 m for the object and 11 m for the observer, so final relative position is +10 m. Relative displacement is 10−(−2) = +12 m; distance is 12 m because relative motion never reverses. Relative velocity is 5−2 = +3 m/s. The initial negative coordinate does not imply negative velocity.