Question

Runner 1 is standing still on a straight running track. Runner 2 passes him, running with a constant speed of \(5.1 \mathrm{~m} / \mathrm{s}\). Just as runner 2 passes, runner 1 accelerates with a constant acceleration of \(0.89 \mathrm{~m} / \mathrm{s}^{2} .\) How far down the track does runner 1 catch up with runner \(2 ?\)

Short answer

The distance down the track when Runner 1 catches up with Runner 2 is approximately 58.44 meters.

Step-by-step solution

Identify the kinematic equations for both runners

Since Runner 1 accelerates constantly and Runner 2 runs at a constant speed, we will write separate kinematic equations for both runners: For Runner 1 (starting from rest with constant acceleration): $$ x_1 = v_{10} t + \frac{1}{2} a_1 t^2 $$ Where \(x_1\) is the distance traveled by Runner 1 at time \(t\), \(v_{10}=0\) is the initial velocity of Runner 1, and \(a_1 = 0.89 ~\mathrm{m}/\mathrm{s}^2\) is the acceleration of Runner 1. For Runner 2 (running at constant speed): $$ x_2 = v_2 t $$ Where \(x_2\) is the distance traveled by Runner 2 at time \(t\) and \(v_2 = 5.1 ~\mathrm{m}/\mathrm{s}\) is the constant velocity of Runner 2.

Set both distances equal to each other

Since we want to find the distance down the track when Runner 1 catches up with Runner 2, we need to find the time \(t\) when \(x_1 = x_2\). Therefore, we can set the equations for \(x_1\) and \(x_2\) equal to each other: $$ v_{10} t + \frac{1}{2} a_1 t^2 = v_2 t $$

Solve for t

Now, substitute the known values of \(v_{10}=0\), \(a_1=0.89 ~\mathrm{m}/\mathrm{s}^2\), and \(v_2 = 5.1 ~\mathrm{m}/\mathrm{s}\) into the equation: $$ 0 + \frac{1}{2}(0.89 ~\mathrm{m}/\mathrm{s}^2) t^2 = (5.1 ~\mathrm{m}/\mathrm{s}) t $$ This equation can be simplified to: $$ 0.445t^2 = 5.1t $$ Now, divide both sides by \(0.445\): $$ t^2 = \frac{5.1t}{0.445} $$ And lastly, divide both sides by \(t\): $$ t = \frac{5.1}{0.445} $$ Calculate the value for \(t\): $$ t \approx 11.46 ~\mathrm{s} $$

Calculate the distance when Runner 1 catches up with Runner 2

Now that we have the time when Runner 1 catches up with Runner 2, we can use either equation for \(x_1\) or \(x_2\) to determine the distance down the track at that moment. We will use the equation for Runner 2, since it is simpler: $$ x_2 = (5.1 ~\mathrm{m}/\mathrm{s})(11.46 ~\mathrm{s}) $$ Calculate the value of \(x_2\): $$ x_2 \approx 58.44 ~\mathrm{m} $$ So, Runner 1 catches up with Runner 2 about \(58.44\) meters down the track.

Key concepts

Constant Acceleration

Understanding constant acceleration is crucial when analyzing the motion of objects, such as Runner 1 in our exercise. Constant acceleration means that the velocity of an object increases by the same amount in every equal time period. For instance, if a car accelerates at a rate of \(3~\mathrm{m/s^2}\), it means that every second, the car's speed will increase by \(3~\mathrm{m/s}\).

In physics, this is described by the kinematic equations, one of which was used to find Runner 1's distance at the catch-up point with Runner 2. With a constant acceleration \(a_1 = 0.89~\mathrm{m/s^2}\), and starting from rest (initial velocity \(v_{10}=0\)), the distance covered over time \(t\) is given by the equation: \[x_1 = \frac{1}{2} a_1 t^2\]. This simplifies the process of solving for time and subsequently the distance, as velocity is not directly influencing the calculations at each moment.

Constant Speed

Constant speed, as seen with Runner 2, implies motion at a steady rate without accelerating or decelerating. It simplifies kinematic analysis because the distance covered is directly proportional to time. Runner 2's speed is \(5.1~\mathrm{m/s}\) and he maintains this speed throughout, showing a linear relationship between distance and time.

This linear relationship is expressed by the equation \[x_2 = v_2 t\], where \(v_2\) is Runner 2's constant velocity and \(t\) is time. Constant speed scenarios are often simpler to analyze compared to those involving acceleration because the rate of change of velocity (i.e., acceleration) is zero.

Distance and Time Relationship

The distance and time relationship is a cornerstone in motion analysis, facilitating the comparison of two moving objects, like our runners. This relationship is fundamental—distance is a result of moving at a certain rate (speed or velocity) for a period of time.

For constant speed, the relationship is linear, as highlighted by \(x_2 = v_2 t\). However, for accelerated motion, like Runner 1’s, the relationship follows a quadratic pattern described by \(x_1 = \frac{1}{2} a_1 t^2\). In our scenario, by setting \(x_1 = x_2\), we equate the distances of both Runner 1 and Runner 2 at the same time, allowing us to solve for \(t\) when their paths align. The result provides the precise instance in time when both runners are at the same spot along the track.

Solving Quadratic Equations

Quadratic equations are fundamental in mathematics and in analyzing scenarios with constant acceleration. They appear in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving these types of equations is crucial, as it was in the step-by-step solution for Runner 1. Here, \(t^2 = \frac{5.1t}{0.445}\), after simplifying, resembles a quadratic equation and is solved by isolating \(t\).

Solving quadratic equations can mean finding the values of \(x\) (or in this case, \(t\)) that make the equation true. There are various methods to solve them, including factoring, completing the square, using the quadratic formula, or graphically. The method chosen often depends on the form of the equation and convenience. In our runners' problem, solving the quadratic equation enabled us to find the time at which Runner 1 caught up to Runner 2.