Question

In Exercises 101 and \(102,\) use the position function \(s(t)=-16 t^{2}+500\) , which gives the height (in feet) of an object that has fallen for \(t\) seconds from a height of 500 feet. The velocity at time \(t=a\) seconds is given by \(\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\) A construction worker drops a full paint can from a height of 500 feet. When will the paint can hit the ground? At what velocity will the paint can impact the ground?

Short answer

The paint can will hit the ground after approximately 5.62 seconds, and it will impact the ground at approximately 179.2 ft/s, directed downwards.

Step-by-step solution

Determine when the can hits the ground

The paint can hits the ground when \(s(t) = 0\). This happens when \(-16t^2 + 500 = 0\). Solving for \(t\), we get \(t = \sqrt{\frac{500}{16}}\).

Compute the velocity at impact

The velocity function is given implicitly by the limit in the problem. However, since velocity is the derivative of the position function with respect to time, we can find the velocity by taking the derivative of the position function. The derivative of \(s(t) = -16t^2 + 500\) with respect to \(t\) is \(v(t) = s'(t) = -32t\). Plug the time \(t\) that we solved in step 1 into the velocity function to find the velocity at impact.

Calculate the exact times and velocities

Solve \(\sqrt{\frac{500}{16}}\) to get the exact time. It equals to approximately 5.62 seconds. The velocity at impact is then given by \(-32 \times 5.62 \approx -179.2\) ft/s. The negative sign indicates that velocity is directed downwards, as expected.

Key concepts

Limit Definition of Velocity

The limit definition of velocity plays a crucial role in understanding motion in calculus. It describes how we can find the velocity of an object at a specific instant, known as the instantaneous velocity. Let's say we have a position function, such as the one given in our exercise, s(t) = -16t^2 + 500. To find velocity using the limit definition, we examine what happens to the position of the object as the time difference approaches zero.

To illustrate this, consider two points in time, t and a, where a is the time of interest and t is another time close to a. We can calculate the average velocity between these two times by the change in position divided by the change in time, (s(a) - s(t)) / (a - t). As t gets closer and closer to a, this average velocity approaches the instantaneous velocity at time a. The exact velocity at time t=a is given by the limit \(lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\). This is the crucial idea behind the limit definition of velocity.

Instantaneous Velocity

Instantaneous velocity is the velocity of an object at a particular instant in time. Unlike average velocity, which measures speed over an interval of time, instantaneous velocity gives us the exact speed and direction of an object at any given moment. In the context of the provided exercise, we're interested in the velocity of a falling paint can at the moment it hits the ground.

To find this, we consider the limit as the time interval becomes infinitesimally small. Instantaneous velocity can be visualized as the slope of the tangent line to the position-time graph at a specific point. This corresponds to the derivative of the position function with respect to time. For example, in the exercise, the instantaneous velocity at time t is found by differentiating the given position function s(t), resulting in the velocity function v(t). By applying this concept, we can determine the velocity of the object at any point during its movement.

Derivative of Position Function

The derivative of a position function represents the velocity function of an object. It provides a mathematical method to find the rate of change of position with respect to time, which is the object's velocity. In our exercise, the position function is s(t) = -16t^2 + 500. By taking the derivative, denoted as s'(t) or v(t), we get the velocity function that tells us the object's velocity at any time t.

Calculus has given us powerful tools, such as derivatives, for analyzing motion. To compute the derivative of -16t^2, we apply the power rule, multiplying the exponent by the coefficient and then subtracting one from the exponent. Consequently, s'(t) = -32t. This simplifies the process of finding instantaneous velocity because we no longer have to deal with limits explicitly, making the derivative a preferable method in most situations.

Solving Quadratic Equations in Calculus

When faced with quadratic equations in calculus, finding solutions is often a necessary step in solving real-world problems, such as predicting the time at which an object will reach the ground when dropped from a certain height. In our exercise, we solve the equation -16t^2 + 500 = 0 to determine when the paint can hits the ground.

The standard form of a quadratic equation is ax^2 + bx + c = 0. To solve it, we can use factoring, completing the square, or the quadratic formula. However, in our scenario, the equation simplifies nicely and we can isolate t by taking the square root after dividing both sides by -16. This approach gives us the precise moment when the paint can will make an impact. It is crucial for students to master various techniques for solving quadratic equations since they frequently appear in calculus problems involving motion and optimization.