In This post you will get answer of Q.
A particle moves along x-axis as
$ x=4,(t-2)+a{{(t-2)}^{2}} $
Which of the flowing is true?
Solution:
Rate of change of displacement is velocity
$ left(v=frac{dx}{dt}right) $
Given, $ x = 4 (t – 2) + a(t – 2)^2 $
using $ frac{d}{dx}x^n=nx^{n-1} $ we have
$ v=frac{dx}{dt}=4+2a(t-2) $
At t = 0, v=4
Acceleration $ a = frac{d^2x}{dt^2} $
$ a=frac{d^2x}{dt^2}=2a $
Solution:
Rate of change of displacement is velocity
$ left(v=frac{dx}{dt}right) $
Given, $ x = 4 (t – 2) + a(t – 2)^2 $
using $ frac{d}{dx}x^n=nx^{n-1} $ we have
$ v=frac{dx}{dt}=4+2a(t-2) $
At t = 0, v=4
Acceleration $ a = frac{d^2x}{dt^2} $
$ a=frac{d^2x}{dt^2}=2a $