Question

The velocity of an object in meters per second is \(v(t)=36-t^{2}, 0 \leq t \leq 6 .\) 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 \(t=3\) seconds is 27 m/s and the acceleration is -6 m/s². The object is slowing down as the velocity and acceleration have opposite signs.

Step-by-step solution

Find the Velocity at t=3

The velocity function is given as \(v(t)=36-t^{2}\). Substitute \(t=3\) into the function, so we get \(v(3)=36-(3)^2=36-9=27\). So the velocity at time \(t=3\) seconds is 27 m/s.

Find the Acceleration at t=3

The acceleration function is the derivative of the velocity function. Therefore, differentiate \(v(t)=36-t^{2}\) with respect to \(t\) to get \(v'(t)=-2t\). Substitute \(t=3\) into the function to get \(v'(3)=-2(3)=-6\). Therefore, the acceleration of the object at \(t=3\) seconds is -6 m/s².

Understand the Physical Meaning

In physics, if the acceleration and velocity have opposite signs it means that the object is slowing down. This is because velocity indicates the speed and direction of an object, whereas acceleration indicates the rate of change of the velocity of the object. So, if they have opposite signs then the object's velocity is decreasing, hence its speed is decreasing too.

Key concepts

Instantaneous Velocity

Instantaneous velocity is a fundamental concept in calculus-based physics which measures the velocity of an object at a specific moment in time. It differs from average velocity which is concerned with the overall change over a time interval. To grasp instantaneous velocity, imagine taking a snapshot of an object's movement so precisely that the time interval is almost zero - that is the essence of instantaneous.In mathematical terms, we determine instantaneous velocity using the derivative of the displacement function with respect to time. For our exercise, we have the velocity function given by the equation v(t) = 36 - t^2. To find the instantaneous velocity at time t = 3 seconds, we simply substitute 3 into the function to get v(3) = 27 m/s. This represents the exact speed and direction of the object at that moment.

Instantaneous Acceleration

Instantaneous acceleration addresses how quickly an object's velocity is changing at a specific instant. Similar to the concept of instantaneous velocity, it's about the rate of change at a particular point in time. To calculate instantaneous acceleration, we need to find the derivative of the velocity function concerning time, which gives us the acceleration function.In the exercise, the velocity function is v(t) = 36 - t^2, and therefore its derivative, or acceleration function, is a(t) = v'(t) = -2t. Plugging in the time t = 3 seconds we get a(3) = -6 m/s², indicating that the object's velocity is decreasing at that moment.

Derivative of Velocity Function

The derivative of the velocity function is crucial in understanding an object's changing motion over time. In the realm of kinematics, the derivative of the velocity function with respect to time yields the acceleration. This relationship is a foundational concept in calculus and helps explain how motion changes.In this particular problem, taking the derivative of the velocity function v(t) = 36 - t^2 gives us v'(t) = -2t. This derived acceleration function tells us that the acceleration is dependent on the time interval, and specifically, it decreases linearly as time increases. At t = 3 seconds, the derivative gives us the instantaneous acceleration, which is -6 m/s².

Kinematics in Calculus

Kinematics in calculus provides a detailed description of motion using the tools of differentiation and integration. Calculus allows us to analyze how quantities like position, velocity, and acceleration change with respect to time. These principles give us not just snapshots of motion at an instant but also how motion evolves over continuous time intervals.The problem at hand showcases a straightforward application of these principles. With the velocity function provided, we first found the instantaneous velocity. Next, we took its derivative to determine the object's instantaneous acceleration. When the acceleration is negative, and the velocity is positive, as in this scenario at t = 3 seconds, this indicates the object is slowing down - a fundamental conclusion drawn using the calculus applied to kinematics.