it is denoted by

v 2 - u 2 = 2as



v = Final Velocity

u = Initial Velocity

s = Distance Travelled

a = acceleration



To solve problems using equations of motion, we should remember that

  • If body starts from rest, its Initial velocity = u = 0
  • If we drop a body from some height, its Initial velocity = u = 0
  • If body stops, its Final velocity = v = 0
  • If body moves with uniform velocity, its Acceleration = a = 0


Q 2 Page 110 - A train is traveling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.

View Answer




Q 5 Page 110 - A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

View Answer



NCERT Question 7 - A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?

View Answer


Example 8.5 - A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

View Answer

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Third Equation of Motion - Derivation 𝑣^2βˆ’π‘’^2 = 2as Derivation We know that Displacement = (πΌπ‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ + πΉπ‘–π‘›π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦)/2 Γ— Time s = ((𝑒 + 𝑣)/2) Γ— t From First equation of motion, v = u + at v βˆ’ u = at t = (𝑣 βˆ’ 𝑒)/π‘Ž Putting value of t in displacement formula s = ((𝑣 + 𝑒)/2) Γ— time s = ((𝑣 + 𝑒)/2) Γ— ((𝑣 βˆ’ 𝑒)/π‘Ž) s = (𝑣^2 βˆ’ 𝑒^2)/2π‘Ž 2as = 𝑣^2βˆ’π‘’^2 𝑣^2βˆ’π‘’^2 = 2as

Ask a doubt
Maninder Singh's photo - Co-founder, Teachoo

Made by

Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo