A particle moving in the xy-plane has velocity v = (2ti + (3-t^2)j) m/s, where t is in s. What is the particle's acceleration vector at t = 4s?

Verified step by step guidance

1. Identify the velocity vector given by the equation $v=(2t\mathbf{i}+(3-t^2)\mathbf{j})$ m/s, where $t$ is the time in seconds.
2. Calculate the acceleration vector, which is the derivative of the velocity vector with respect to time $t$. Differentiate each component of the velocity vector separately.
3. Differentiate the x-component of the velocity vector $2t$ with respect to $t$ to find the x-component of the acceleration vector.
4. Differentiate the y-component of the velocity vector $3-t^2$ with respect to $t$ to find the y-component of the acceleration vector.
5. Evaluate the acceleration components at $t=4$ seconds to find the acceleration vector at that specific time.

Key Concepts

Velocity

Velocity is a vector quantity that describes the rate of change of an object's position with respect to time. It has both magnitude and direction, and in this case, it is given as a function of time in the xy-plane. Understanding velocity is crucial for determining how the position of the particle changes over time.

Acceleration

Acceleration is the rate of change of velocity with respect to time. It is also a vector quantity and can be calculated by taking the derivative of the velocity vector. In this problem, finding the acceleration at a specific time involves differentiating the velocity function provided.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. In physics, it is used to determine rates of change, such as how velocity changes over time to find acceleration. For the given velocity function, applying differentiation will yield the acceleration vector at any time t.