Question

If position of a particle at instant t is given by x=t^5+2t+3 find acceleration of the particle

Answer 1

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

Answer 2

acceleration = 20 * t^3
thanks...