Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings ?

Q: What is the answer to NDA (II) 2023 GAT Q131?

D. Correct — D, the upward-bending curve. The catch is in the axes: all four printed graphs plot **distance against time**, while the equation quoted, v = u + at, is a statement about velocity. So the question is asking what a distance–time graph looks like for the motion that v = u + at describes. v = u + at is the motion of a body under **uniform acceleration**. Integrating it gives the distance travelled, s = ut + ½at², which is quadratic in t — a curve, not a line. Because the question insists all quantities are non-zero, the acceleration a cannot vanish, so the ½at² term is always present and the graph can never straighten out. That disposes of (a), (b) and (c) at once: every one of them is a straight line, and a straight distance–time line means constant velocity, which is precisely the a = 0 case the question excludes. Only (d) is curved, and it bends upwards, as a positive acceleration requires.

(a)A distance–time graph that is a straight line through the origin
(b)A distance–time graph that is a straight line starting above the origin
(c)A distance–time graph that is a straight line of negative slope
(d)A distance–time graph that is a curve starting above the origin and bending upwards

Answer

Why

Correct — D, the upward-bending curve. The catch is in the axes: all four printed graphs plot distance against time, while the equation quoted, v = u + at, is a statement about velocity. So the question is asking what a distance–time graph looks like for the motion that v = u + at describes.

v = u + at is the motion of a body under uniform acceleration. Integrating it gives the distance travelled, s = ut + ½at², which is quadratic in t — a curve, not a line. Because the question insists all quantities are non-zero, the acceleration a cannot vanish, so the ½at² term is always present and the graph can never straighten out.

That disposes of (a), (b) and (c) at once: every one of them is a straight line, and a straight distance–time line means constant velocity, which is precisely the a = 0 case the question excludes. Only (d) is curved, and it bends upwards, as a positive acceleration requires.

Why the others are wrong

(a)
A distance–time graph that is a straight line through the origin — A straight distance–time line means the body covers equal distances in equal times — constant velocity, a = 0. The question rules that out by stating all quantities are non-zero.
(b)
A distance–time graph that is a straight line starting above the origin — Still a straight line, so still constant velocity. The starting offset only means the body began at some distance from the reference point; it does nothing about the acceleration.
(c)
A distance–time graph that is a straight line of negative slope — A negative slope means the body is moving back towards the origin at a steady speed — again uniform velocity, and again no acceleration.

Concept

The three equations of motion for uniform acceleration are v = u + at, s = ut + ½at² and v² = u² + 2as. On a velocity–time graph, v = u + at is a straight line of slope a and intercept u. On a distance–time graph the same motion appears as a parabola, because distance depends on the square of the time. The gradient of a distance–time graph is the velocity; the gradient of a velocity–time graph is the acceleration.

Read the axis labels before the equation. A candidate who sees v = u + at and pictures the familiar sloping straight line is remembering a velocity–time graph and will pick (a) or (b) — but these axes say distance. The phrase 'all quantities are non-zero' is the second signpost: it exists to force a ≠ 0, which is what makes the curve compulsory. Curvature on a distance–time graph is the visual signature of acceleration.

Key facts

v = u + at describes uniformly accelerated motion.
The matching distance relation is s = ut + ½at², quadratic in t.
A straight distance–time line means constant velocity, a = 0.
Curvature on a distance–time graph indicates acceleration.
On a velocity–time graph the same equation would be a straight line of slope a.

Which axes are these?

v = u + at plotted as velocity vs time→Straight line, slope a, intercept u — but that is NOT what is printed

The same motion as distance vs time→s = ut + ½at² — a curve

Straight distance–time line→Constant velocity, a = 0 — excluded by 'all quantities non-zero'

The graphs are distance–time, so uniform acceleration must appear as a curve — option (d).

Study next

The three equations of uniformly accelerated motionDistance–time versus velocity–time graphsGradient and area interpretations on motion graphsRecognising acceleration from curvature

Common traps

Reading the equation and picturing the wrong pair of axes.
Ignoring the condition that all quantities are non-zero, which forces a ≠ 0.
Assuming a straight line always represents uniform acceleration.

NDA quotes an equation and prints graphs on axes that may not match it — identify the axes first, convert the equation to the right relation, then pick the shape.

Related PYQs

NDA_GAT_2018_I_Q542018Open & attempt →

An object is moving with uniform acceleration a. Its initial velocity is u and after time t its velocity is v. The equation of its motion is v = u + at. The velocity (along y-axis) time (along x-axis) graph shall be a straight line

(a) passing through origin
(b) with x-intercept u
(c) with y-intercept u
(d) with slope u

Answer(c) with y-intercept u

The same equation on the axes it actually belongs to — there the v-versus-t plot is a straight line whose y-intercept is the initial velocity u. Seeing that makes clear why the distance–time version here must curve instead.

NDA_GAT_2022_II_Q972022Open & attempt →

What is the nature of velocity-time graph for a car moving with uniform acceleration?

(a) Parabola
(b) Logarithmic
(c) Straight line
(d) Exponential

Answer(c) Straight line

Asks the shape of the velocity–time graph under uniform acceleration — a straight line, which is exactly the shape candidates wrongly transfer onto the distance–time axes in this item.

Practice

practice — not a real PYQ
The distance–time graph of a body moving with uniform velocity is
- (a)a curve bending upwards
- (b)a straight line
- (c)a horizontal line
- (d)a parabola
Answer(b) a straight line — equal distances in equal times give a constant gradient.
practice — not a real PYQ
On a velocity–time graph, the equation v = u + at appears as a straight line whose y-intercept is
- (a)the acceleration a
- (b)the initial velocity u
- (c)the distance s
- (d)zero
Answer(b) the initial velocity u — at t = 0, v = u.