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it is denoted by

v 2 - u 2 = 2as

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Where

v = Final Velocity

u = Initial Velocity

s = Distance Travelled

a = acceleration

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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

Questions

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.

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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?

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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?

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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.

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Transcript

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

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CA Maninder Singh is a Chartered Accountant for the past 12 years and a teacher from the past 16 years. He teaches Science, Accounts and English at Teachoo