Q.According to v-t graph, why the acceleration of a body along OA-graph is greater than along OB-graph?

The rate of change of velocity of an object with respect to time is called its acceleration.

Velocity can change in magnitude (speeding up/slowing down), direction, or both. Acceleration is defined as:

a = dv/dt = (change in velocity)/(time taken)

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With uniform acceleration the velocity may either increase or decrease.

Uniform acceleration means the velocity changes by equal amounts in equal times. If the acceleration is in the same direction as the velocity, the speed increases; if it is opposite (retardation), the speed decreases.

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Constant acceleration means a straight-line v–t graph with constant slope — that's graph (A).

Reasoning. For constant acceleration $a$, the equation of motion is $v = u + at$, which is linear in $t$ (a straight line with slope $a$ and intercept $u$).

Checking each sketch:

• (A) Straight line from the origin with constant positive slope — velocity increases uniformly with time, exactly $v=at$ (starting from rest). This matches constant acceleration.
• (B) Horizontal line — velocity doesn't change with time, so acceleration is zero, not constant nonzero acceleration.
• (C) A vertical line — implies velocity changing instantaneously with no time elapsed, i.e. infinite acceleration, which is unphysical. …

Constant slope of a v–t graph ⇒ uniform acceleration.

Acceleration $a = \dfrac{dv}{dt}$ = slope of the velocity–time graph.

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$s = kt^2$ differentiates to a constant acceleration $a = 2k$.

Given $s \propto t^2$, write $s = k t^2$ for some constant $k$.

Velocity: $v = \dfrac{ds}{dt} = 2kt$

Acceleration: $a = \dfrac{dv}{dt} = 2k$

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A displacement-time graph for uniform acceleration is a concave-up curve that gets steeper with time, since s depends on t^2 -- graph (c).

For motion with uniform (constant) acceleration a, the displacement-time relation is

s = ut + (1/2)a t^2

This is a quadratic function of time. Its slope, ds/dt = u + at, is the instantaneous velocity, which increases steadily with time for positive a. A steadily increasing slope means the curve bends upward more and more sharply as t increases -- this is exactly a concave-up curve that starts at the origin and curves increasingly steeply, matching graph (c).

By contrast:

(a) a straight line through the origin with constant slope describes UNIFORM VELOCITY (zero acceleration), not uniform acceleration. …