Exam level5 past questions

Equations of Motion and Uniform Acceleration

Learn the kinematic equations for uniformly accelerated motion, including applications to free fall, retardation, and calculation of distance in specific time intervals.

Why this shows up in the exam

You need to apply these equations to solve problems involving objects moving with constant acceleration, a frequent NEET topic.

How NEET tests this

Numerical · 4 QsDirect recallApplication

Learn the idea

Uniform acceleration means the acceleration stays constant, so the velocity changes linearly and the distance travelled equals the average velocity multiplied by time.

🧠 Memory hook: Distance = average speed × time → average speed is simply (initial + final)/2 for constant acceleration.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • v = u + a t
  • s = u t + ½ a t²
  • s = (u + v) t / 2
  • a is constant → velocity‑time graph is a straight line
  • Convert km/h to m/s by multiplying by 5/18

How to approach it

  1. 1List the given quantities and convert them to SI units
  2. 2Find the missing acceleration or velocity using v = u + a t if needed
  3. 3Choose the SUVAT formula that contains the known variables and the required distance
  4. 4Compute s and check the unit

Worked example — watch it click

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of:

  • A)20 m
  • ✅400 m
  • C)1440 m
  • D)2880 m

The concept behind this problem

The worked example asks for the distance when a car starts from rest and reaches a known speed in a known time, exactly the situation where s = (u+v) t /2 or s = ½ a t² applies.

Step by step

  1. 1Final velocity = 144 km/h = 144×(5/18) = 40 m/s.
  2. 2Initial velocity u = 0, t = 20 s.
  3. 3Using s = ut + (1/2)at²: First find a = (v-u)/t = 40/20 = 2 m/s².
  4. 4Then s = 0 + (1/2)(2)(20²) = 400 m.
  5. 5Or use s = (u+v)t/2 = (0+40)×20/2 = 400 m.

Watch out

Students often multiply the final speed (40 m/s) by the time (20 s) and get 800 m, forgetting that the speed was not constant.

Common slip-ups that cost marks

  • •Using final speed directly as distance = v·t (ignores that speed is not constant)
  • •Forgetting to convert km/h to m/s before using the equations
  • •Mixing up sign of acceleration when the motion is retardation

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5NEET 1997

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s. Then it covers a distance of:

Push further

More challenging

4 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 4

A particle moves such that its velocity is given by v(x) = P/x + Q, where P and Q are constants and x is the position. Find the acceleration of the particle as a function of x.