Physics - Kinematics Question with Solution | TestHub

$$a=v\frac{dv}{dx}$$

So, we have the differential equation:

$$v\frac{dv}{dx}=3x^2-2x+6$$

Now, we separate the variables and integrate both sides:

$$\int vdv=\int (3x^2-2x+6)dx$$

$$\frac{v^2}{2}=x^3-x^2+6x+C$$

We are given an initial condition: when the particle is at the origin ($x=0$), its speed is $6m/s$. We can use this to find the constant $C$.

Substitute $x=0$ and $v=6$ into the equation:

$$\frac{(6{)}^2}{2}=(0{)}^3-(0{)}^2+6(0)+C$$

$$\frac{36}{2}=0+C$$

$$18=C$$

Now, substitute the value of $C$ back into the equation:

$$\frac{v^2}{2}=x^3-x^2+6x+18$$

We need to find the velocity of the particle when it is at position $x=6\mathrm{m}$. Substitute $x=6$ into the equation:

$$\frac{v^2}{2}=(6{)}^3-(6{)}^2+6(6)+18$$

$$\frac{v^2}{2}=234$$

$$v^2=468$$

$$v=6\sqrt{13}m/s$$