Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Two objects move along a coordinate line. At the end of \(t\) seconds their directed distances from the origin, in feet, are given by \(s_{1}=4 t-3 t^{2}\) and \(s_{2}=t^{2}-2 t\), respectively. (a) When do they have the same velocity? (b) When do they have the same speed? (c) When do they have the same position?

Step-by-Step Solution

Answer

(a) \(t = \frac{3}{4}\) sec, (b) \(t = \frac{1}{2}, \frac{3}{4}\) sec, (c) \(t = 0, \frac{3}{2}\) sec.

Step 1: Express velocity equations

To find when the two objects have the same velocity, we first need to express their velocity functions by differentiating their position functions. The velocity of the first object is given by the derivative of \( s_1 = 4t - 3t^2 \), resulting in \( v_1 = \frac{d}{dt}(4t - 3t^2) = 4 - 6t \). Similarly, the velocity of the second object is given by the derivative of \( s_2 = t^2 - 2t \), resulting in \( v_2 = \frac{d}{dt}(t^2 - 2t) = 2t - 2 \).

Step 2: Solve for the same velocity

To determine when the velocities are equal, set the velocity equations equal to each other: \( 4 - 6t = 2t - 2 \). Simplify and solve for \( t \): 1. Add \(6t\) to both sides: \(4 = 8t - 2\). 2. Add \(2\) to both sides: \(6 = 8t\). 3. Divide by \(8\): \(t = \frac{3}{4}\) seconds.

Step 3: Express speed equations

Speed is the absolute value of velocity, so we take absolute values of \(v_1\) and \(v_2\): - \(|v_1| = |4 - 6t| \). - \(|v_2| = |2t - 2| \). Equate these to find when speeds are the same.

Step 4: Solve for the same speed

Solve \(|4 - 6t| = |2t - 2| \). Consider two cases for solving absolute value equations:1. **Case 1:** \(4 - 6t = 2t - 2\). - This is already solved in Step 2: \(t = \frac{3}{4}\). 2. **Case 2:** \(4 - 6t = -(2t - 2)\). Simplifying: - \(4 - 6t = -2t + 2\), add \(2t\) to both sides: \(4 - 4t = 2\). - Subtract 4 from both sides: \(-4t = -2\). - Divide by \(-4\): \(t = \frac{1}{2}\). So, speeds are the same at \(t = \frac{3}{4}\) seconds and \(t = \frac{1}{2}\) seconds.

Step 5: Equate position equations

To determine when the objects are at the same position, set their position functions equal: \(4t - 3t^2 = t^2 - 2t\). Simplify and solve for \( t \): 1. Rearrange: \(4t - 3t^2 - t^2 + 2t = 0\). 2. Combine like terms: \(-4t^2 + 6t = 0\). 3. Factor: \(-2t(2t - 3) = 0\).

Step 6: Solve for the same position

Solve the factored equation: 1. \(-2t = 0\) gives \(t = 0\). 2. \(2t - 3 = 0\) solves to \(2t = 3\), or \(t = \frac{3}{2}\). Thus, the objects are in the same position when \(t = 0\) and \(t = \frac{3}{2}\) seconds.

Key Concepts

Velocity Equations

Velocity is a crucial concept in calculus motion problems, as it describes the rate at which an object's position changes over time. To find the velocity, we differentiate the position function with respect to time. In our exercise, we have two position functions for two objects.- For the first object, the position equation is given by \( s_1 = 4t - 3t^2 \). To find the velocity equation, differentiate to get \( v_1 = \frac{d}{dt}(4t - 3t^2) = 4 - 6t \).- For the second object, the position equation is \( s_2 = t^2 - 2t \). Differentiating this gives the velocity equation \( v_2 = \frac{d}{dt}(t^2 - 2t) = 2t - 2 \).Knowing the velocity equations is important because it allows you to determine how quickly each object is moving in a given direction at any moment. This foundational step is necessary before comparing velocities to find when they are the same.

Speed Equations

Speed is the non-directional measure of how fast an object is moving. It is the absolute value of velocity, and considering it as such is crucial when direction doesn't matter, like in many real-life scenarios.- The speed of the first object is represented by \(|v_1| = |4 - 6t|\).- The speed of the second object is \(|v_2| = |2t - 2|\).To find when the two objects have the same speed, we need to consider both cases of absolute values and solve: - Case 1: Set the expressions equal, as in the velocity problem, which we've already solved to give \( t = \frac{3}{4} \).- Case 2: Consider both possible ranges for when the velocities' signs differ, leading to another value of \( t = \frac{1}{2} \).This makes it clear that the objects can have the same speed, even when their velocities differ, due to the speed being an absolute measure.

Position Equations

Position equations describe an object's location along a coordinate line at any given time. In solving for when two objects are at the same position, we set their position functions equal to each other. This involves some algebraic manipulation and often results in solving quadratic equations.- For our objects, equate the position functions: \(4t - 3t^2 = t^2 - 2t\).- Simplify to \(-4t^2 + 6t = 0\), which can be factored into \(-2t(2t - 3) = 0\).Solving this equation provides values of \( t \) that indicate when both objects are at the same position:- From \(-2t = 0\), the solution is \( t = 0 \).- From \(2t - 3 = 0\), solve for \( t = \frac{3}{2} \).Understanding position equations and their solutions helps you pinpoint exact moments in time where two objects coincide in position along their paths. This is essential for analyzing scenarios in physical motion and predicting future positions.