Q.v-t curve for an object with uniform acceleration is given below.
Using this graph derive the relation, x = v0 t + (1/2) a t^2 and v^2 = v0^2 + 2 a x.

The three kinematic equations $v=u+at$, $s=ut+\tfrac12at^2$, $v^2=u^2+2as$ are derived assuming $a$ is constant — that restriction (stated by R) is exactly why they take this particular simple form (stated by A).

Assertion (A): For uniformly accelerated motion, integrating $a=\dfrac{dv}{dt}=\text{constant}$ gives $v=u+at$; integrating $v=\dfrac{ds}{dt}$ gives $s=ut+\tfrac12at^2$; eliminating $t$ between these gives $v^2=u^2+2as$. This is true — these are exactly the standard three equations of uniformly accelerated motion.

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The statement is TRUE: stopping distance s ∝ u².

For a vehicle moving with initial velocity u and decelerating uniformly at rate a (due to braking) until it stops (final velocity v = 0), the third equation of motion gives:

v² = u² − 2as

0 = u² − 2as

s = u² / (2a)

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s = ½at^2 = ½(2)(10)^2 = 100 m. Answer (C).

The body starts from rest, so initial velocity u = 0.

Using the equation of motion s = ut + ½at^2:

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Since time t is not given, use the third equation of motion, v^2 = u^2 + 2as, which relates v, u, s, and a without t.

The three standard equations of motion for uniform acceleration are:

1. v = u + at (needs t)
2. s = ut + (1/2)at^2 (needs t)
3. v^2 = u^2 + 2as (needs u, a, s — no t) …

With u = 0 and constant a = 2 m/s² for t = 5 s, the car covers 25 m.

For motion with constant acceleration starting from rest, the displacement is given by the second equation of motion:

$s = ut + \frac{1}{2}at^2$

Here u = 0 (starts from rest), a = 2 m/s², t = 5 s.

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The first equation of motion is v = u + at, matching option (e).

For a body moving with uniform (constant) acceleration a, starting with initial velocity u, its velocity v after time t is given by the first equation of motion: …

The second equation of motion is S = ut + (1/2)at^2, matching option (c).

For a body moving with uniform acceleration a starting with initial velocity u, the displacement S covered in time t is given by the second equation of motion: …

Using x = (1/2)at^2 and the given final velocity v = an, the average velocity over the last 2 seconds works out to v(n-1)/n.

Body starts from rest (u = 0) with uniform acceleration a. Its velocity after n seconds is

v = a n => a = v/n

Displacement as a function of time: x(t) = (1/2) a t^2

Displacement in the last 2 seconds (from t = n-2 to t = n):

Δx = x(n) - x(n-2) = (1/2)a[n^2 - (n-2)^2] = (1/2)a[(n-(n-2))(n+(n-2))] = (1/2)a(2)(2n-2) = 2a(n-1)

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First equation of motion, v = u + at, directly gives the final velocity.

Given: initial velocity u = 20 m/s, acceleration a = 2 m/s^2, time t = 10 s.

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