Question

For the motion of an object along the $x-axis$, the velocity $v$ depends on the displacement $x$ as $v=3x^2-2x$, then what is the acceleration at $x=2m$.

Answer 1

Step 1: Given data

$v=3x^2-2x$, $x=2m$

Step 2: Formula used

$a=\frac{∆v}{∆t}[a=accleration,\frac{∆v}{∆t}=velocitychangeovertime]$

Step 3: Calculating acceleration at $x=2m$.

Acceleration describes the rate at which an object's velocity varies. If an object's velocity changes frequently, it is still accelerating. Acceleration is a vector quantity.

$$\begin{gathered} v=3x^2-2x \\ a=\frac{∆v}{∆t} \\ a=\frac{dv}{dt} \\ or \\ a=\frac{dv}{dx}\times \frac{dx}{dt} \\ a=v\frac{dv}{dx} \\ a=(3x^2-2x)(6x-2) \end{gathered}$$

We need to find acceleration at $x=2m$

After using the value of x,

$$\begin{gathered} a=[(3\times 2^2)-2\times 2]\left[\right(6\times 2)-2] \\ a=(12-4)(12-2) \\ a=8\times 10 \\ a=80m/s^2 \end{gathered}$$

Therefore, acceleration is $80m/s^2$.