- #1
brotherbobby
- 699
- 163
- Homework Statement
- A particle moves in the ##x-y## plane with constant acceleration ##a_0## directed along the negative y–axis. The equation of motion of the particle has the form ##y=px-qx^2## where ##p## and ##q## are positive constants. Find the velocity of the particle at the origin of coordinates.
- Relevant Equations
- 1. If ##\dot x = v_x=\dfrac{dx}{dt}## is the ##x## component of velocity and ##\dot y = v_y=\dfrac{dy}{dt}## the ##y## component of velocity, then the (scalar) velocity or speed at a point is given by ##v = \left( \dot x^2+\dot y^2\right)^{\frac{1}{2}}##.
2. Acceleration components ##a_x=\ddot x## and ##a_y=\ddot y##.
3. Acceleration can also be expressed as a function of position, for instance, ##a_x=v_x\dfrac{dv_x}{dx}##.
Problem statement : I copy and paste the problem as it appears in the text down below. I have only changed the symbol of the given acceleration from ##a\rightarrow a_0##, owing to its constancy.
Attempt : I must admit that I could proceed very little.
Given ##a_0=\dfrac{d^2y}{dt^2}\Rightarrow \dot y = a_0t+b##. But the constant ##b## is unknown.
Likewise, it is also given that ##\dfrac{d^2x}{dt^2}=0\Rightarrow \dot x = c##, but the constant ##c## is unknown.
Differentiating the path of the particle ##y = px-qx^2## on both sides with respect to time, ##\dot y = p\dot x-2qx\dot x\Rightarrow a_0t+b = pc-2qcx##.
This equation can yield the position ##x## as a function of the unknowns ##b,c##.
Substituting for ##x## in the equation of the curve can yield ##y## again as a function of the same unknowns.
I can't see how to progress from here.
Request : A suggestion as to how to proceed to solve the problem will be welcome.
Attempt : I must admit that I could proceed very little.
Given ##a_0=\dfrac{d^2y}{dt^2}\Rightarrow \dot y = a_0t+b##. But the constant ##b## is unknown.
Likewise, it is also given that ##\dfrac{d^2x}{dt^2}=0\Rightarrow \dot x = c##, but the constant ##c## is unknown.
Differentiating the path of the particle ##y = px-qx^2## on both sides with respect to time, ##\dot y = p\dot x-2qx\dot x\Rightarrow a_0t+b = pc-2qcx##.
This equation can yield the position ##x## as a function of the unknowns ##b,c##.
Substituting for ##x## in the equation of the curve can yield ##y## again as a function of the same unknowns.
I can't see how to progress from here.
Request : A suggestion as to how to proceed to solve the problem will be welcome.