An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant accel

Numerical · Q16

Q. An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 s. What is the velocity of the aeroplane at the take off?

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Concept understanding — Equations of Motion

When acceleration is constant, three equations tie together the initial velocity u, final velocity v, acceleration a, time t and displacement s. They are the workhorses of JEE Main kinematics — the whole skill is picking the right one and keeping the signs honest.

The three equations (uniform acceleration only):

v = u + at — no s
s = ut + ½at² — no v
v² = u² + 2as — no t (use this whenever time is neither given nor asked)

Plus the distance in the nth second: sₙ = u + a(n − ½) — note this is a distance covered during one second, not a total distance (a favourite trap). And the average velocity under uniform acceleration is (u + v)/2.

1 — Signs are everything. Fix a positive direction first. A body slowing down has a opposite to v (negative if v is positive). Stopping distance comes from v² = u² + 2as with v = 0: s = u²/(2a) — so it scales as u² (double the speed → four times the stopping distance). Total stopping distance with reaction time = u·t_react + u²/(2a).

2 — Motion under gravity is just constant acceleration with a = g (take g = 10 m/s² unless told otherwise). Key results, with up taken positive for a body thrown up at speed u:

maximum height H = u²/2g; time to the top = u/g; time up = time down; total flight = 2u/g;
the speed on returning to the launch level equals u (same magnitude, opposite direction);
at the highest point the velocity is zero but the acceleration is still g downward. For a body dropped from rest: h = ½gt², v = gt, v² = 2gh, and the distances in successive seconds are in the ratio 1 : 3 : 5 : 7 … (Galileo's odd-number rule; cumulative distances go as t², i.e. 1 : 4 : 9).

3 — Thrown from a height / released from a moving carrier. Set the net displacement to −h (ground below the start) and solve the quadratic −h = ut − ½gt². A body released from a rising balloon keeps the balloon's upward velocity as its own initial velocity (it first goes up, then falls); from a descending lift it starts downward. Thrown up vs thrown down from the same height give the same landing speed (v² = u² + 2gh) but different times.

4 — Two-body and multi-stage problems. For two bodies under gravity, write each one's position on a common clock and set them equal to find where/when they meet; since both share the same g, their relative acceleration is zero, so their separation changes at the constant relative speed. For multi-stage motion (accelerate → cruise → brake), carry the end velocity of each phase into the next. A body passing a height twice on the way up and down has the two times as the roots of h = ut − ½gt² (t₁ + t₂ = 2u/g, t₁t₂ = 2h/g).

How this concept is examined. JEE Main asks for a missing kinematic variable by direct substitution, an nth-second distance, a stopping distance (mind the u² scaling), a free-fall time/speed, a max height or time of flight, a meeting point of two bodies, or a Galileo-ratio result. The equations are few; the marks reward choosing the time-free equation when appropriate and never dropping a sign.

Equations of motion are introduced in the NCERT Class 11 Physics chapter on Motion in a Straight Line, and are among the most searched board-revision topics under queries like 'equations of motion formula class 11 physics' and 'kinematics important questions.' Because nearly every JEE Main and NEET mechanics numerical starts from these three relations, mastering the sign conventions here pays off well beyond the CBSE syllabus.

Given: $u=0$ , $s=500$ m, $t=30$ s. Using the second equation of motion, $s=ut+\tfrac12at^2$ , with $u=0$ : $500 = \tfrac12 a(30)^2 = 450a$ , so $a = \dfrac{500}{450} = \dfrac{10}{9} \approx 1.11$ m/s². Then using the first equation, $v=u+at = 0 + \dfrac{10}{9}\times 30 = \dfrac{100}{3} \approx 33.33$ m/s. Converting to km/hr: $33.33 \times 3.6 = 120$ km/hr. [!ANSWER] v = 120 km/hr (33.3 m/s)

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Two statements are given below, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes below. Assertion (A): The motion of an object moving with uniform acceleration can be expressed by three equations: v = u + at, s = ut + (1/2)at², v² = u² + 2as Reason (R): These equations do not apply to non-uniform or variable acceleration.$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) and (R) both are false.
›Reveal solutionSolution

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.

Reason (R): Every step of that derivation used $a=\text{constant}$ pulled out of the integral. If acceleration varies with time, these same integrals give different (and generally more complicated) results, so the fixed three-equation forms genuinely break down for non-uniform acceleration. This is also true, and it is precisely the reason the assertion's three equations look the way they do — the constancy of $a$ is the single assumption baked into their derivation, so R directly explains why A's equations are valid only in the uniform-acceleration case A describes.

✓Final answer
$(a) Both (A) and (R) are true and (R) is the correct explanation of (A).$
CBSE 2026Set ANNUAL1 mark
$Q. Write true or false: The stopping distance of a moving vehicle is proportional to the square of its initial velocity.$
›Reveal solutionSolution

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)

Since a (the braking deceleration) is roughly constant for a given road/vehicle condition, the stopping distance s is directly proportional to u² — doubling the initial speed quadruples the stopping distance. This is exactly why speed limits are so important for road safety.

✓Final answer
$True — stopping distance s = u²/(2a), so it is proportional to the square of the initial velocity.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If the acceleration of a body moving from rest is 2 m/s^2, then the distance travelled by the body in 10 sec will be$
(a) 10 m
(b) 50 m
(c) 100 m
(d) 20 m
›Reveal solutionSolution

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:

s = (0)(10) + ½(2)(10)^2 = ½ x 2 x 100 = 100 m.

✓Final answer
$(C) 100 m.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which equation of motion can be used to find the final velocity when initial velocity, displacement and acceleration are known?$
(a) v = u + at
(b) s = ut + (1/2)at^2
(c) v = u + 2as
(d) v^2 = u^2 + 2as
›Reveal solutionSolution
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:

v = u + at (needs t)

s = ut + (1/2)at^2 (needs t)

v^2 = u^2 + 2as (needs u, a, s — no t)

Since the problem gives initial velocity (u), displacement (s), and acceleration (a) but not time, only the third equation lets you solve directly for v.

✓Final answer
$(d) v^2 = u^2 + 2as.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. A car accelerated from rest at 2 m/s². How far will it travel in 5 second ?$
(a) 10 m
(b) 25 m
(c) 50 m
(d) 100 m
›Reveal solutionSolution

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.

$s = 0 \times 5 + \frac{1}{2}(2)(5)^2 = \frac{1}{2}(2)(25) = 25\ \text{m}$

✓Final answer
$The car travels 25 m in 5 seconds. Option (b) 25 m.$
CBSE 2024Set ANNUAL1 mark
$Q. Match the term with its correct dimensional formula / equation from this pool (each option is used exactly once):$
(a) [M0LT-2],
(b) [MLT-2],
(c) S = ut + (1/2)at^2,
(d) [M-1L3T-2],
(e) v = u + at,
(f) [MLT-1]. Which option correctly matches 'First equation of motion'?
›Reveal solutionSolution

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:

v = u + at.

This directly matches option (e).

✓Final answer
$First equation of motion matches with (e) v = u + at.$
CBSE 2024Set ANNUAL1 mark
$Q. Match the term with its correct dimensional formula / equation from this pool (each option is used exactly once):$
(a) [M0LT-2],
(b) [MLT-2],
(c) S = ut + (1/2)at^2,
(d) [M-1L3T-2],
(e) v = u + at,
(f) [MLT-1]. Which option correctly matches 'Second equation of motion'?
›Reveal solutionSolution

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:

S = ut + (1/2) a t^2.

This directly matches option (c).

✓Final answer
$Second equation of motion matches with (c) S = ut + (1/2)at^2.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. A body initially at rest is moving with uniform acceleration a. Its velocity after n seconds is v. The velocity of the body in last 2 s is$
(1) 2v(n-1)/n
(2) v(n-1)/n
(3) v(n+1)/n
(4) 2v(n+1)/n
›Reveal solutionSolution

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)

The velocity 'in the last 2 s' asked here is this displacement divided by the 2-second interval (average velocity over that interval):

v_(last 2s) = Δx / 2 = a(n-1)

Substituting a = v/n:

v_(last 2s) = v(n-1)/n

✓Final answer
$(2) v(n-1)/n.$
CBSE 2022Set TERM11 markMCQ
$Q. A body starts at a velocity of 20 m/s. If its acceleration is 2 m/s^2, then velocity of the body after 10 second is$
(1) 40 m/s
(2) 20 m/s
(3) 30 m/s
(4) 10 m/s
›Reveal solutionSolution

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.

Using v = u + at:

v = 20 + (2)(10) = 20 + 20 = 40 m/s.

✓Final answer
$(1) 40 m/s.$

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Q.An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 s. What is the velocity of the aeroplane at the take off?

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