A particle's position is r⃗=(c t^2-2 d t^3) +(2 c t^2-d t^3) where c and d are positive constan

step 1 All right, so we have a position vector and we want to know when our time is greater than zero,

A particle's position is r⃗=(c t^2-2 d t^3) +(2 c t^2-d t^3)...

A particle's position is $\vec{r}=\left(c t^{2}-2 d t^{3}\right) \hat{\imath}+\left(2 c t^{2}-d t^{3}\right) \hat{\jmath}$
where $c$ and $d$ are positive constants. Find expressions for times $t>0$ when the particle is moving in (a) the $x$ -direction and (b) the $y$ -direction.

Key Concepts

Vector Differentiation

This concept involves taking the derivative of a vector-valued function with respect to time to obtain the velocity vector. Differentiation is used to determine how each individual component of the position vector changes over time, which is fundamental to the study of kinematics in physics.

Velocity Vector

The velocity vector represents the rate of change of the position vector and indicates both the speed and the direction of a particle's motion. By differentiating each component of the position vector, one obtains the respective components of the velocity vector, which are then used to analyze the motion.

Directional Motion Conditions

For a particle to move only along a specific axis (such as the x-axis or y-axis), its velocity component in the perpendicular direction must vanish. This means setting the irrelevant component of the velocity vector (either the y-component for x-direction motion or the x-component for y-direction motion) to zero, while ensuring that the remaining component is nonzero.

Polynomial Equation Solving

After setting the appropriate velocity component to zero to satisfy directional motion conditions, one often obtains a polynomial equation in time. Solving these equations requires techniques from algebra such as factoring and identifying physically meaningful solutions (e.g., considering only positive time values in the context of the problem).

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