Question

Acceleration The velocity of an object in meters per second is $$ v(t)=36-t^{2} $$ for \(0 \leq t \leq 6 .\) Find the velocity and acceleration of the object when \(t=3 .\) What can be said about the speed of the object when the velocity and acceleration have opposite signs?

Short answer

The velocity of the object at time \( t=3 \) is 27 m/s, and acceleration is \( -6 \: m/s^2\). When the acceleration and velocity have opposite signs, the object is decelerating.

Step-by-step solution

Calculate the velocity at t=3

By inserting \( t = 3 \) into the velocity function \( v(t)=36-t^{2} \), the velocity at time \( t=3 \) can be calculated: \( v(3) =36-3^{2}= 27 \: m/s \)

Calculate the derivative of the velocity function to get the acceleration

The acceleration of an object is essentially the rate at which its velocity is changing and therefore is the derivative of the velocity function. This can be found by applying the power rule of differentiation, giving the acceleration function as \( a(t) = -2t \).

Calculate the acceleration at t=3

By inserting \( t = 3 \) into the acceleration function \( a(t)=-2t\), the acceleration at time \( t=3 \) can be calculated: \( a(3) =-2\times3=-6 \: m/s^{2} \).

Analyze the sign of velocity and acceleration

When the velocity and acceleration have opposite signs, it means the object is decelerating. Here, at time \( t = 3 \), the velocity is positive and the acceleration is negative, which implies the object is decelerating at that point in time.

Key concepts

Velocity Function

In the exercise, we faced the velocity function given by \( v(t)=36-t^2 \). The velocity function describes the speed and direction of a moving object over time. It shows how fast an object is traveling in a specific direction. In this case, the velocity is a function of time, which varies as time changes. The expression \( 36-t^2 \) points out that the object's speed decreases as time progresses due to the negative \( t^2 \) term.

• Initial Velocity: When \( t = 0 \), \( v(0) = 36 \) meters per second.
• Decreasing Trend: The term \(-t^2\) causes the velocity to decrease, indicating that the object slows down as time goes on.
• Specific Example: At \( t = 3 \), inserting into the velocity function gives \( v(3) = 27 \) meters per second, meaning the object is still moving forward but slowing down.

The velocity function is crucial as it tells us not just how fast something moves, but also how that speed changes over time.

Derivative of a Function

The derivative of a function in calculus is a powerful tool for understanding how functions change. In the context of this exercise, we derived the velocity function to find the acceleration—how quickly the velocity itself changes with time. Differentiating a function involves calculating its rate of change, often represented as the derivative.

• Finding Derivative: To get the acceleration function from the velocity function \( v(t) = 36 - t^2 \), we apply the power rule. The derivative is \( a(t) = -2t \).
• Power Rule: This rule states that for any term \( t^n \), the derivative is \( nt^{(n-1)} \). Here, \( -t^2 \) differentiates to \( -2t \).
• Interpreting Derivative: The derivative tells us how the velocity—our primary interest here—is changing over time, making it key to understanding acceleration.

Knowing how to find and interpret the derivative allows us to anticipate and analyze changes in motion, such as speeding up or slowing down.

Deceleration Concept

Understanding deceleration is essential when discussing motion, especially when velocity and acceleration have opposing signs. In the provided example, at \( t = 3 \), the velocity is \( 27 \) meters per second (a positive value) while the acceleration is \(-6 \) meters per second squared (a negative value). This condition — where velocity is positive and acceleration is negative — indicates deceleration.

• Opposing Signs: Deceleration occurs when acceleration opposes the current direction of motion. Here, a positive velocity and negative acceleration suggests the object is slowing down.
• Practical Example: Imagine driving a car where you're still moving forward (positive velocity), but pressing the brakes (negative acceleration) to gradually decrease speed.
• Physics of Deceleration: This is indicative of a force acting in the opposite direction of motion, reducing the velocity over time.

Understanding deceleration helps in predicting how the object's speed changes, preparing us to calculate the stopping time or distance required. By analyzing velocity and acceleration, students can determine when objects slow down and the effects of forces acting on them.