Question

General head start problem Suppose object A is located at \(s=0\) at time \(t=0\) and starts moving along the s-axis with a velocity given by \(v(t)=2 a t,\) where \(a>0 .\) Object \(B\) is located at \(s=c>0\) at \(t=0\) and starts moving along the \(s\) -axis with a constant velocity given by \(V(t)=b>0 .\) Show that A always overtakes \(\mathrm{B}\) at time $$t=\frac{b+\sqrt{b^{2}+4 a c}}{2 a}$$

Short answer

Question: Determine the validity of the given solution for the time when object A overtakes object B, with the given position functions \(s_A(t) = at^2\) and \(s_B(t) = bt + c\). The given solution for the time \(t\) is: $$t = \frac{b + \sqrt{b^2 + 4ac}}{2a}$$ Answer: The given solution for the time \(t\) when object A overtakes object B is valid.

Step-by-step solution

Find the position functions of both objects A and B

To find the position functions for objects A and B, we need to integrate their velocity functions. For A: The given velocity function for A is \(v(t) = 2at\). We will integrate this with respect to time (t) to find the position function, \(s_A(t)\). $$s_A(t) = \int v(t) dt = \int 2at dt = at^2 + C_A$$ Since A is at \(s=0\) at \(t=0\), we can find the constant \(C_A\): $$0 = 0^2 + C_A \Rightarrow C_A = 0$$ Hence, the position function for object A is $$s_A(t) = at^2$$ For B: The given constant velocity function for B is \(V(t)=b > 0\). We will integrate this with respect to time (t) to find the position function, \(s_B(t)\). $$s_B(t) = \int V(t) dt = \int b dt = bt + C_B$$ Since B is at \(s=c>0\) at \(t=0\), we can find the constant \(C_B\): $$c = 0 + C_B \Rightarrow C_B = c$$ Hence, the position function for object B is $$s_B(t) = bt + c$$

Solve for the time when the objects have the same position

We will now set the position functions of objects A and B equal to each other and solve for time t when they are in the same position. $$at^2 = bt + c$$ Rearrange the equation to get a quadratic equation in the form of \(at^2 - bt - c = 0\): $$at^2 - bt - c = 0$$

Apply the quadratic formula to solve for t

This quadratic equation can be solved using the quadratic formula: $$t = \frac{-b \pm \sqrt{(-b)^2 - 4ac}}{2a}$$ Plugging the coefficients from our equation (\(a\), -b, -c) into the quadratic formula gives: $$t = \frac{b \pm \sqrt{(-(-b))^2 - 4(a)(-c)}}{2(a)}$$ Now we will simplify the expression: $$t = \frac{b \pm \sqrt{b^2 + 4ac}}{2a}$$ Assuming that A overtakes B, the overtaking time would be at the largest positive root, which means we select the positive square root: $$t = \frac{b + \sqrt{b^2 + 4ac}}{2a}$$ And this matches the given solution.

Key concepts

Position Functions

In the realm of calculus and motion, position functions serve as a mathematical description of an object's location relative to a reference point over time. They can be thought of as the output of cumulatively adding up (integrating) the velocity of an object with respect to time.

For instance, if an object starts from a known position and moves according to a specific velocity function, its position function can be found by integrating the velocity function with respect to time. This is essential in problems involving motion since knowing an object's position at any given time allows us to understand and predict its trajectory.

In the given exercise, Object A starts at the origin with a position function represented by s_A(t) = at², which indicates a changing velocity since the velocity increases with time. On the other hand, Object B has a constant velocity, leading to a linear position function, s_B(t) = bt + c. By comparing these functions, we can predict and analyze the movement interaction between the two objects—namely, when and where Object A overtakes Object B.

Quadratic Formula

The quadratic formula is a powerful tool in algebra, providing the solution(s) for any quadratic equation of the form ax² + bx + c = 0. The formula is x = (-b \(pm\) \sqrt{b² - 4ac})/(2a).

Application in Motion Problems

When we apply this formula to motion problems, particularly those involving position functions like in our exercise, it allows us to solve for the time when two moving objects will intersect or overlap in their paths. After setting up the quadratic equation by equating the two position functions, we can use the formula to find the precise moment of this interaction.

As in our exercise, given the equation at² - bt - c = 0, we use the quadratic formula and choose the positive square root for time because we're interested in the time after the start when A overtakes B. This is because time cannot be negative in this context, and we are looking for the first instance of overtaking.

Definite Integrals

In calculus, definite integrals are employed to compute the accumulated quantity, such as area under a curve, distance, or—in our context—the total change in position or displacement over a period. Essentially, a definite integral takes a function and evaluates it between two bounds, providing a numerical result.

For motion problems, integrating the velocity function of an object over a certain time interval gives us the object's displacement. If the velocity is constant, the definite integral is straightforward, essentially multiplying velocity by the time elapsed. If velocity changes with time, as it does for Object A in the exercise, the integration process captures this variation to give an accurate representation of the position over time.

Understanding definite integrals allows students to not only determine the position at a specific time but also to calculate the object's average velocity, total distance traveled, and other essential characteristics of its motion.