Interpreting Motion Graphs
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Direct answer
A position-time (x-t) graph encodes velocity as its slope: a flat line means rest, a straight sloping line means uniform velocity (v = Δx/Δt), and a curved line means accelerated motion, with the curvature's sign giving the direction of acceleration. A velocity-time (v-t) graph is richer — its slope gives acceleration (a = Δv/Δt) and the area under it gives displacement, taken negative where the curve dips below the time axis. An acceleration-time graph gives change in velocity as its area. Most NEET questions on this topic are exercises in reading these three quantities off a sketch, so the slope-area rules, applied with sign discipline, settle nearly every question in under a minute.
What you must remember
- Slope rules: slope of x-t = velocity; slope of v-t = acceleration; slope of a-t is not a standard quantity (its area is, and equals Δv).
- Area rules: area under v-t = displacement (not distance — parts below the axis subtract); area under a-t = change in velocity Δv.
- Shape dictionary: uniformly accelerated motion gives a parabola in x-t (opening along the acceleration direction), a straight inclined line in v-t, and a horizontal line in a-t.
- Reversal condition: a body reverses direction only where v changes sign, i.e. where the v-t curve crosses the time axis or the x-t curve has a peak (zero slope) — never where a = 0 momentarily by itself.
- Average versus instantaneous: the chord slope of an x-t graph between two points is average velocity; the tangent slope at a point is instantaneous velocity. NEET loves asking which is which on a curved x-t sketch.
- Standard value hook: with g = 9.8 m/s^2 (NCERT convention; take 10 only if the paper gifts it), a free-fall v-t graph has slope 9.8 downward, and the area after t seconds works out to 4.9 t^2 metres of fall.
- Distance trap: for a v-t graph that goes positive then negative with equal areas, net displacement is zero but the distance travelled is twice one area — read the question's last word carefully.
Walking through a typical NEET sketch
Suppose a v-t graph rises linearly from 0 to 20 m/s in 4 s, stays at 20 m/s for 4 s, then falls linearly to −10 m/s over the next 6 s. The three slopes give the accelerations: 20/4 = 5 m/s^2, then 0, then (−10 − 20)/6 = −5 m/s^2. For displacement, add the areas by inspection: a triangle of base 4 s and height 20 (40 m), a rectangle 4 × 20 (80 m), then a trapezium from t = 8 s to 14 s — the part above the axis is 20 × 4 × ½ = 40 m, and the triangle below the axis from t = 12 s to 14 s is ½ × 2 × 10 = 10 m with a negative sign. Total displacement = 40 + 80 + 40 − 10 = 150 m, while distance = 160 m. Notice how the zero-crossing at t = 12 s is the pivot: that is where the body stops and turns back, and every quantity after it flips sign. Working a full graph this way once is worth memorising ten formulas, because the exam rarely deviates from this triangle-rectangle-trapezium toolkit.
How the exam frames graphs
The recurring trap is graphical language that sounds precise but is not: "the slope of the graph is negative" is meaningless until you name which graph — negative v-t slope means retardation only while velocity is positive, and acceleration (not retardation) if velocity is already negative. A second favourite asks for the point of maximum speed on a curved x-t graph; students point to the steepest visible segment, which is right, but then misidentify it because the curve keeps steepening beyond the drawn window. NCERT Class 11, Chapter 3 builds the whole discussion around three named examples — a car at rest, one at uniform velocity, one accelerating — and NEET options are frequently the same three shapes shuffled, so anchor your answers to that NCERT trio rather than to exotic curves.
Frequently asked questions
What does the area under a velocity-time graph represent?
It represents displacement; the portions below the time axis contribute negatively, so the total area with signs gives displacement while the sum of magnitudes gives distance travelled.
When is a position-time graph a straight line?
Only for rest (horizontal) or uniform velocity (inclined); any acceleration, constant or changing, bends the x-t graph, becoming a parabola when acceleration is uniform.
How do you find where a body reverses its direction from a graph?
Locate the instant velocity changes sign — the v-t curve crossing the time axis, equivalently a maximum or minimum (zero slope) point on the x-t curve.
Can a body have zero velocity yet non-zero acceleration?
Yes: at the topmost point of vertical throw or at the turning point of an oscillation, velocity is momentarily zero while acceleration remains 9.8 m/s^2 or a = −ω^2 x respectively.
What is the difference between average and instantaneous velocity on a graph?
Average velocity is the slope of the chord joining two points on the x-t graph; instantaneous velocity is the slope of the tangent at a single point, and the two coincide only for straight-line (uniform velocity) graphs.