On a velocity-time (v-t) graph, the slope of the line at any point equals the acceleration at that instant, because a = change in velocity / change in time = rise / run. A steep slope means large acceleration; a flat (horizontal) line means zero acceleration; a downward slope means negative acceleration (retardation). Memory hook: "v-t slope = a" — the graph goes UP one step, slope UP means speeding up.
On a velocity-time graph the slope equals acceleration: a rising line (green) means positive acceleration, a flat line (blue) means zero acceleration, and a falling line (red) means negative acceleration (retardation).
Your doubts, answered
Does the slope of a velocity-time graph give acceleration or displacement?
Slope gives acceleration, not displacement. Slope means rise/run, which is (change in velocity)/(change in time) = a. Displacement is a different operation: it is the AREA under the v-t graph, not the slope. Rule to remember: on a v-t graph, slope = acceleration, area = displacement.
What does a flat (horizontal) velocity-time graph mean?
A horizontal line means velocity is not changing. Slope = 0, so acceleration = 0. The body moves with constant (uniform) velocity. It is still moving (velocity is not zero unless the line sits on the v = 0 axis), it just is not speeding up or slowing down.
How do I find acceleration from a v-t graph in a numerical?
Pick two clear points on the line, read their velocities and times. Then a = (v2 - v1) / (t2 - t1). Example: if v goes from 4 m/s at t = 1 s to 10 m/s at t = 3 s, then a = (10 - 4)/(3 - 1) = 6/2 = 3 m/s^2. Units of the slope are (m/s) / s = m/s^2, which are the correct units of acceleration.
Why is a downward-sloping v-t line negative acceleration?
If the line goes down as time increases, velocity is decreasing, so change in velocity is negative. Slope = negative rise / positive run = negative. This is negative acceleration, also called retardation or deceleration. For a body slowing down, the v-t graph slopes down toward the time axis.
How is slope of v-t different from area under v-t?
They answer two different questions. Slope of the v-t line tells you HOW FAST the velocity is changing, which is acceleration (m/s^2). Area between the line and the time axis tells you HOW FAR the body moved, which is displacement (m). NEET often puts both ideas in one question, so keep them separate: slope = a, area = s.
How do I match a v-t graph to the correct a-t graph?
Read the slope of the v-t graph piece by piece. Where the v-t line rises, a is a positive constant. Where the v-t line is flat, a = 0. Where the v-t line falls, a is a negative constant. Draw those constant values on the a-t graph as horizontal steps. A straight sloping v-t line always gives a constant (horizontal) a-t line.
⚠️ The NEET trap ✗ Reading the acceleration off the HEIGHT of the v-t graph (thinking high velocity means high acceleration). ✓ Acceleration is the SLOPE of the v-t line, not its height. A point can be high up (large velocity) yet have zero slope (zero acceleration), like a fast car moving at constant speed. Always look at how steep the line is, not how high it sits. 🧠 Height of the v-t line = velocity. Steepness (slope) of the line = acceleration. Two different things.
Real NEET questions
NEET 2024
The velocity (v)-time (t) graph of a body moving in a straight line first rises, then stays flat, then falls. Which acceleration (a)-time (t) graph best represents the motion?
A · a constant positive throughout
B · a > 0, then a = 0, then a < 0 ✓
C · a constant negative throughout
D · a increases linearly
Solution: Acceleration is the slope of the v-t graph, read piece by piece. Stage 1: the v-t line rises, so slope is positive and constant, a > 0. Stage 2: the v-t line is flat, so slope = 0, a = 0. Stage 3: the v-t line falls, so slope is negative and constant, a < 0. The correct a-t graph is a positive step, then zero, then a negative step. Answer: (B).
NEET 2026
A ball is thrown vertically upward and falls back to the thrower. Taking upward as positive, which velocity (v)-time (t) graph correctly represents its motion?
A · v stays positive throughout
B · horizontal straight line
C · a straight line of constant negative slope, crossing v = 0 ✓
D · a curved (parabolic) line
Solution: The only force is gravity, so acceleration is constant at -g the whole time (up and down). Since a = slope of the v-t graph, the v-t graph must be a single STRAIGHT line with constant negative slope. Velocity starts at +u, decreases through 0 at the highest point, then becomes negative (-u) as the ball falls. So the line crosses the v = 0 axis once. Answer: (C).
Solved Motion In A Straight Line NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
In one line, what does the slope of a v-t graph give?
It gives the acceleration at that instant, because slope = (change in velocity)/(change in time) = a.
What are the units of the slope of a v-t graph?
The rise is in m/s and the run is in s, so the slope has units (m/s)/s = m/s^2, which are exactly the units of acceleration.
If the v-t line is a curve, how do I get the acceleration?
Draw the tangent to the curve at that point and find the slope of that tangent. The slope of the tangent equals the instantaneous acceleration, since a = dv/dt.
Can acceleration be positive while velocity is negative?
Yes. Sign of velocity is the position of the line (above or below the axis); sign of acceleration is the slope. A line can be below the axis (negative velocity) but rising toward it (positive slope, positive acceleration).
Why is this a common NEET question?
NEET regularly asks you to match a v-t graph to the correct a-t graph, or to compare slope with area. Knowing 'slope = acceleration, area = displacement' lets you answer these fast without any calculation.
