Distance and relative motion

01 / UNDERSTAND & APPLY

Distance and displacement are different targets

Distance is total path length travelled. Displacement is the change from starting to ending position, with direction. Average speed is total distance divided by elapsed time; average velocity is displacement divided by elapsed time. A round trip can have positive average speed and zero average velocity.

Convert units before combining rates and lengths. One kilometre is 1000 metres and one hour 3600 seconds, so 1 km/h=5/18 m/s. Reverse the conversion by multiplying 18/5. Track whether minutes or hours appear in a graph’s horizontal axis.

Velocity on a one-dimensional coordinate line can be negative because it includes direction. Speed is its nonnegative magnitude. A negative position coordinate is a location relative to the chosen origin, not a negative distance travelled.

02 / UNDERSTAND & APPLY

Constant speed and elapsed time

For constant speed v over time t, distance d=vt. Solve t=d/v only with positive v for a positive-distance journey. If speed varies, calculate separate segments or use total distance and total time; a single instantaneous speed is not automatically the journey’s average.

Stops contribute elapsed time while adding no distance. If the question asks average moving speed, it may exclude stops, but a start-to-finish average normally includes them. State the convention rather than omitting waiting time for convenience.

03 / UNDERSTAND & APPLY

Combine distances and times, not speeds alone

For several segments, add distances and elapsed times independently, then divide. Equal-time segments permit a simple arithmetic mean of speeds; equal-distance segments generally do not because the slower segment lasts longer.

Include every stated stop in an overall elapsed-time average. A speed estimate should also fit the journey: without stops, the average lies between segment speeds; adding stops can move it below the lowest moving speed.

04 / UNDERSTAND & APPLY

Measure how quickly the gap changes

For two objects moving toward one another along the same line, their closing speed is the sum of their positive speeds. For one catching another in the same direction, it is the faster speed minus the slower speed. These rules assume constant speeds and the stated directions over the interval.

Meeting time is initial gap divided by positive closing speed. If the same-direction relative speed is zero, a nonzero initial gap persists. If the pursuer is slower, it does not catch the leader under an unchanged model; a negative algebraic time is not a future meeting.

Check each object’s distance using its own speed and the same elapsed interval. In an approach problem, those distances should sum to the original separation.

05 / UNDERSTAND & APPLY

What must completely pass what?

When a train passes a stationary point, its front must travel one train length from first arrival until the rear clears the point. To pass a platform completely, the front travels train length plus platform length. The relevant distance is not always the visible obstacle’s length alone.

For two trains to cross completely, use the sum of their lengths and the relative speed. Opposing trains use the speed sum; same-direction overtaking uses the positive difference. Define the starting moment as their leading ends meeting for an opposing crossing, or the faster front reaching the slower rear for a same-direction full pass.

Convert km/h to m/s before dividing lengths in metres. A unit mismatch can create a plausible-looking but wildly incorrect time.

06 / UNDERSTAND & APPLY

Boat speed relative to water and ground

In a simple straight-stream model, downstream ground speed equals still-water boat speed plus current speed. Upstream ground speed equals still-water speed minus current speed. The model assumes constant speeds and the boat travelling directly along the stream.

Positive upstream progress requires still-water speed greater than current speed. If they are equal, the boat holds position relative to the bank while pointing upstream; it cannot cover a positive upstream distance. A stronger current moves it downstream under the unchanged model.

For a return journey, calculate each leg’s time separately. Equal distances upstream and downstream do not produce equal times, so the average ground speed is not simply the still-water speed.

07 / UNDERSTAND & APPLY

Read the vertical variable before taking a slope

On a position-time graph, slope is signed velocity: rise in position divided by elapsed time. A horizontal segment is rest at a fixed position; a negative slope indicates motion toward decreasing coordinates. It does not mean negative speed.

A cumulative-distance graph measures path already travelled. It cannot decrease. Its slope is speed, so return travel makes cumulative distance rise even while position falls. For the example below, cumulative-distance points would be(0,0),(5,400),(8,400),(12,800), with final slope+100 m/min; the position graph instead returns to 0 with slope−100 m/min.

For a piecewise straight graph, calculate each segment separately and add the absolute position changes for total distance in this one-dimensional model. Do not confuse the final vertical coordinate with the accumulated path.

08 / UNDERSTAND & APPLY

Use one time origin

A delayed start creates an initial gap by the time the pursuer begins. You may solve from that later start or use a common clock variable, but do not compare distances measured over different durations as though they cover the same interval.

Check the final meeting positions and both elapsed times. A route may also contain a stop or change of direction, requiring stages rather than one relative-speed formula. State the time reference in the answer: after the first start and after the second start are different quantities.

These calculations use ideal constant-rate models. Their purpose is reasoning with supplied quantities and graphs, not forecasting real traffic or travel conditions.