Question

Displacement x of a particle is related to time t as x =at+bt^2-ct^3 , where a, b, c are constants of the motion.The velocity of the particle when its acceleration is zero is given by:

Answer 1

Answer:

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Answer 2

The velocity of the particle when its acceleration is zero is given by $a+\dfrac{b^2}{3c}$.

Explanation:

The displacement of a particle is given by :

$x=at+bt^2-ct^3$

Where

a,b and c are constants of motion

Velocity,

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(at+bt^2-ct^3)}{dt}\\\\v=a+2bt-3ct^2$ ......(1)

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(a+2bt-3ct^2)}{dt}\\\\a=2b-6ct$

When a is 0,

$2b-6ct=0\\\\t=\dfrac{b}{3c}$

Put the value of t in equation (1) as :

$v=a+\dfrac{2b^2}{3c}-\dfrac{b^2}{3c}\\\\v=a+\dfrac{b^2}{3c}$

So, the velocity of the particle when its acceleration is zero is given by $a+\dfrac{b^2}{3c}$. Hence, this is the required solution.

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