Question

For moving particle, the relation between time and position is given by t = Ax^2+Bx. Where A and B
are constants. Find out the acceleration of the particle as a function of velocity.

Answer 1

Answer on Question #43809 – Physics - Mechanics | Kinematics | Dynamics

For moving particle, the relation between time and position is given by $t = A x^2 + B x$. Where A and B are constants. Find out the acceleration of the particle as a function of velocity.

Solution:

Relation between time and position:

$$t = A x^2 + B x$$

The derivative of a function representing the position of a particle along a line at time $t$ is the instantaneous velocity at that time. The derivative of the velocity, which is the second derivative of the position function, represents the instantaneous acceleration of the particle at time $t$.

Derivative of the both parts of the initial equation:

$$\begin{array}{l} dt = \frac{d(A x^2 + B x)}{d x} \\ dt = (2 A x + B) d x \\ (2 A x + B) \frac{d x}{d t} = 1 \\ \end{array}$$

The velocity of the particle is:

$$v = \frac{d x}{d t} = \frac{1}{2 A x + B} = (2 A x + B)^{-1}$$

The acceleration of the particle is:

$$a = v' = \frac{d v}{d t} = \frac{d (2 A x + B)^{-1}}{d t} = - (2 A x + B)^{-2} \cdot 2 A \frac{d x}{d t} = - v^2 \cdot 2 A v = - 2 A v^3$$

Answer: acceleration of the particle is $a = -2 A v^3$

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