Question

A particle moves in straight line. Its position is given by \(\mathrm{x}=2+5 \mathrm{t}-3 \mathrm{t}^{2}\). Find the ratio of initial velocity and initial acceleration. \((\mathrm{A})+(5 / 6)\) (B) \(-(5 / 6)\) (C) \((6 / 5)\) (D) \(-(6 / 5)\)

Short answer

The ratio of initial velocity and initial acceleration is \(-\dfrac{5}{6}\).

Step-by-step solution

Find the velocity function

To find the velocity function, take the first derivative of the position function with respect to time: \(v(t) = \dfrac{d}{dt}(2 + 5t - 3t^2)\) \(v(t) = 5 - 6t\)

Find the initial velocity

To find the initial velocity, substitute t=0 into the velocity function: \(v(0) = 5 - 6(0) = 5\)

Find the acceleration function

To find the acceleration function, take the second derivative of the position function with respect to time: \(a(t) = \dfrac{d^2}{dt^2}(2 + 5t - 3t^2)\) \(a(t) = -6\) Since acceleration is constant, the initial acceleration is also -6.

Find the ratio of initial velocity and initial acceleration

To find the ratio of initial velocity and initial acceleration, divide the initial velocity by the initial acceleration: Ratio = \(\dfrac{v(0)}{a(0)} = \dfrac{5}{-6}\) So the ratio is \(-\dfrac{5}{6}\). The correct answer is (B) \(-\dfrac{5}{6}\).

Key concepts

Understanding Velocity

In kinematics, velocity describes the rate at which an object's position changes over time. It has both magnitude and direction, making it a vector quantity.

• Position Function: This is the equation that provides the location of an object at a given time. In our problem, it is given by \(x(t) = 2 + 5t - 3t^2\).
• First Derivative: To find velocity \(v(t)\), we differentiate the position function with respect to time. This gives us \(v(t) = 5 - 6t\).
• Initial Velocity: By setting \(t = 0\), we can find the object's velocity at the beginning of its motion: \(v(0) = 5\). This tells us how fast the particle was moving initially.

Velocity is crucial when predicting where an object will be in the future, and it provides key insights into the object's motion.

Acceleration: The Change in Velocity

Acceleration refers to how quickly an object's velocity changes with time. It is also a vector, indicating both the rate of change of speed and its direction.

• Finding Acceleration: To find acceleration, we take the second derivative of the position function, yielding \(a(t) = -6\). This constant value indicates the uniform rate of change in velocity.
• Initial Acceleration: Since the acceleration is constant, the initial acceleration is the same at all times, including at \(t = 0\), so \(a(0) = -6\).
• Acceleration informs us about changes in velocity. A negative acceleration, as seen here, suggests that the velocity is decreasing as time progresses.

Understanding acceleration allows us to comprehend how forces acting on the particle influence its motion.

Derivatives and Their Role in Motion

Derivatives are fundamental in kinematics for describing and understanding motion. They allow us to transition from position to velocity, and from velocity to acceleration.

• First Derivative: This derivative of the position function with respect to time gives us the velocity function. It tells us the instantaneous rate of change of position.
• Second Derivative: The subsequent derivative of the velocity function leads to the acceleration function. It reveals how the velocity changes over time.
• Application: In this exercise, derivatives help find both velocity \(v(t) = 5 - 6t\) and acceleration \(a(t) = -6\), providing insight into the particle's motion characteristics.

Mastering derivatives in kinematics equips students with the tools to predict and interpret the behavior of moving objects effectively.