a particle confined to motion along x axis moves with a constant accelaration fro x=2m to x=8m during the 2.5secs time interval. the velocity of particle at x=8m is 2.8m/s. what is the accelaration during this time interval?

Question

Swapnil Saxena · Answer

Make use of Newtons 1, 3 equation

t=2.5, s=6, v=2.8

v= u+a(2.5) --(i)

v^2= u^2+2a(6)---(iii)

Using equation I and II we get (v+u )=24/5= 4.8

When v= 2.8 the the initial velocity of the particle is 2m/s

Usin i equation

Acceleartion is 0.8/2.5 = 8/25 = 0.32m/s^2

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a particle confined to motion along x axis moves with a constant accelaration fro x=2m to x=8m during the 2.5secs time interval. the velocity of particle at x=8m is 2.8m/s. what is the accelaration during this time interval?

a particle confined to motion along x axis moves with a constant accelaration fro x=2m to x=8m during the 2.5secs time interval. the velocity of particle at x=8m is 2.8m/s. what is the accelaration during this time interval?

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