if position of a particle at instant t is given by x= t3.find the acceleration of the particle..

derieve the relation for uniformly accelerated motion along the straight line. s= ut + 1/2 at2where the symbols have their usual meaing....

Consider a body having initial velocity u,If it is subjected to an acceleration 'a' so that after time 't' its velocity becomes v.

acceleration=Change in velocity/time taken

a=(v-u)/t

v=u+at

Suppose the distance travelled by the above body in time t=s

we have average velocity=(Initial velocity+Final velocity)/2

(1) 

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Sure i will do it tomorrow

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Position is given by x = t^3

We know that the second derivative of distance x gives acceleration.

So differentiate it twice, first you get velocity, then you get acceleration

First we get v = dx/dt = 3t^2

then we get acceleration , dv/dt = 6t

Make sure you know differentiation formula d(x^n) / dt = n x^(n - 1)

For the second part of your question,

We will derive the relation using average velocity

Average velocity = Total distance covered / total time taken

Let the object started with initial velocity u, after accelerating for time t, it's final velocity becomes v

then we can write acceleration a = (v - u)/t

v = u + at ----equation (1)

Also we know that the avergae velocity during this interval is the average of u and v

that is, (u + v)/2

Total distance covered = average velocity x time

s = [(u + v)/2] x t

substitute the value of v from equation 1

s = [(u + u + at)/2] x t

s = (2u + at)*t/2

s = [2ut + at^2]/2

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

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