Which feature of a position-time graph gives an object's velocity at a moment?
A floor-cleaning robot rolls 8 m out from its dock at a steady 2 m/s, pauses for 2 s, then swings around and comes home faster and faster, reaching the dock 10 s after it left. That is one trip. It can be drawn three ways, and the three drawings must agree.
Keep the robot's trip in hand. Stage 1: out at 2 m/s for 4 s, reaching 8 m. Stage 2: a 2 s pause. Stage 3: a speeding-up return that arrives back at the dock at 10 s, moving at 4 m/s the other way, so v = −4 m/s.
The robot's whole trip, told three ways on one shared clock: position on top, velocity in the middle, acceleration at the bottom. Every stage boundary lines up down the stack.
All three graphs describe the same trip, so they must agree at every moment.
Rolling out, 0 s to 4 s: the robot covers 8 m at a steady 2 m/s. Its position-time graph climbs as a straight line from 0 m to 8 m, slope 2 m/s. So its velocity-time graph holds a flat level at 2 m/s, because the slope of a position-time graph at a moment is the velocity at that moment, sign included.
Pausing, 4 s to 6 s: the robot sits at 8 m. Its position-time line runs flat, slope zero, so its velocity-time level drops to 0 m/s.
Returning, 6 s to 10 s: the robot speeds up toward the dock. Its position-time graph is a curve that steepens steadily downward, ending at the marked slope of −4 m/s. A steepening curve means the velocity keeps changing, so the velocity-time graph there is a straight ramp from 0 down to −4 m/s.
Now build the bottom graph from the middle one. Each straight velocity-time piece's slope becomes a flat acceleration level, because the slope of a velocity-time graph at a moment is the acceleration at that moment, sign included. The robot's flat velocity levels give 0 m/s². For the return ramp:
slope = (−4 − 0) ÷ 4 = −1 m/s²
So the return stage is a flat acceleration level at −1 m/s².
Check the seams. The robot changes what it is doing at 4 s and at 6 s. Wherever a graph changes behavior, the change sits at one of those clock times: the boundaries line up down the stack. A graph can also show no change at a seam. The acceleration stays at 0 m/s² across the 4 s seam, because both stages hold a steady velocity.
That is the whole method, for any multi-stage trip. Build the velocity-time graph from the position-time graph stage by stage. Then build the acceleration-time graph from the velocity-time graph. Keep every stage boundary lined up on the shared time axis.
A shuttle bus's position-time graph: out from 0 m to 6 m at a steady pace over 2 s, flat at 6 m until 4 s, then a steady return to 0 m by 7 s.
Draw the bus's matching velocity-time and acceleration-time graphs on paper, stacked below the position-time graph with the time axes aligned. Then continue to compare your drawing with the model answer.
The model answer. Check your drawing: the stage slopes are 6 ÷ 2 = 3 m/s, then 0 m/s, then (0 − 6) ÷ 3 = −2 m/s, so the velocity levels are 3, 0, and −2 m/s; every velocity level is flat, so the acceleration is 0 m/s² for the whole 7 s.
The figure shows a tugboat's position-time graph — out, pause, return, all at steady paces — and four candidate pairs of velocity-time and acceleration-time graphs, A to D. Which panel shows the pair that matches the motion?
A courier cart's position-time graph: out from 0 m to 8 m at a steady pace over 2 s, flat at 8 m until 5 s, then a return whose curve steepens steadily, reaching 4 m at 9 s with a marked final slope of −2 m/s.
Draw the cart's matching velocity-time and acceleration-time graphs on paper, time axes aligned. Then continue to compare your drawing with the model answer.
The model answer. Check your drawing: stage 1's slope is 8 ÷ 2 = 4 m/s and stage 2 gives 0 m/s; the steepening curve starts from rest, so the velocity ramps straight from 0 down to the marked −2 m/s at 9 s; the ramp's slope is (−2 − 0) ÷ 4 = −0.5 m/s², so stage 3's acceleration level sits at −0.5 m/s².
A mail cart's position-time graph: out at a steady pace, a pause, then a steady return.
On paper, draw the mail cart's matching velocity-time and acceleration-time graphs directly below the position-time graph, all three time axes aligned. Label every axis with quantity and unit and mark each level's value. Then continue to compare your drawing with the model answer.
Check your drawing: the velocity levels are 6 ÷ 3 = 2 m/s, then 0 m/s, then (0 − 6) ÷ 2 = −3 m/s; every velocity level is flat, so the acceleration-time graph sits at 0 m/s² for the whole trip; every change of behavior sits at 3 s or 5 s, and the acceleration graph shows no change at either seam; every axis is labeled with its quantity and unit.
A rail cart's position-time graph: out at a steady pace, a pause, then a return whose curve steepens steadily to a marked final slope of −4 m/s.
On paper, draw the rail cart's matching velocity-time and acceleration-time graphs directly below the position-time graph, all three time axes aligned. Label every axis with quantity and unit and mark each level's and ramp endpoint's value. Then continue to compare your drawing with the model answer.
Check your drawing: the velocity has a level at 6 ÷ 2 = 3 m/s, a level at 0 m/s during the pause, then one straight ramp from 0 down to the marked −4 m/s; the return is a ramp, not a level, because the steepening curve means the velocity keeps changing; the ramp's slope is (−4 − 0) ÷ 2 = −2 m/s², so the acceleration is 0 m/s² until 3 s and then a level at −2 m/s²; every change of behavior sits at 2 s or 3 s, and the acceleration graph changes only at 3 s.
Your turn
An airport cart's position-time graph with four stages: a steady roll, a stretch whose curve steepens steadily to a marked slope of 4 m/s, a pause, and a steady return.
On paper, draw the airport cart's matching velocity-time and acceleration-time graphs directly below the position-time graph, all three time axes aligned. Label every axis with quantity and unit and mark each level's and ramp endpoint's value. Then continue to compare your drawing with the model answer.
Check your drawing: the velocity runs 4 ÷ 2 = 2 m/s, then a straight ramp from 2 m/s up to the marked 4 m/s, then 0 m/s, then (0 − 10) ÷ 2 = −5 m/s; the steepening stage's ramp starts from the 2 m/s the cart already has, not from zero; the acceleration is 0 m/s² everywhere except 2 s to 4 s, where the ramp's slope gives (4 − 2) ÷ 2 = 1 m/s²; the fast steady return still has a = 0 m/s², because its velocity level is flat.
The figure shows a harbor shuttle's position-time graph — out, pause, return, all at steady paces — and four candidate pairs of velocity-time and acceleration-time graphs, A to D. Which panel shows the pair that matches the motion?
A luggage cart's position-time graph: out at a steady pace, a pause, then a return whose curve steepens steadily to a marked final slope of −8 m/s as it reaches the origin.
On paper, draw the luggage cart's matching velocity-time and acceleration-time graphs directly below the position-time graph, all three time axes aligned. Label every axis with quantity and unit and mark each level's and ramp endpoint's value. Then continue to compare your drawing with the model answer.
Check your drawing: the velocity has a level at 8 ÷ 2 = 4 m/s, a level at 0 m/s during the pause, then one straight ramp from 0 down to the marked −8 m/s; the return stage is a ramp below the axis, not a level, because the cart keeps gaining speed in the negative direction; the ramp's slope is (−8 − 0) ÷ 2 = −4 m/s², so the acceleration is 0 m/s² until 4 s and then a level at −4 m/s²; every change of behavior sits at 2 s or 4 s, and the acceleration graph changes only at 4 s.