Problem 16 - Applied Mathematics

Question

In a motor race, a car \(\mathrm{A}\) is \(1 \mathrm{~km}\) from the finishing post, and is travelling at \(35 \mathrm{~m}\) per second with a uniform acceleration of \(\frac{3}{3} \mathrm{~m}\) per \(\mathrm{sec}^{2}\). At the same instant a second car \(\mathrm{B}\) is \(200 \mathrm{~m}\) behind \(\mathrm{A}\) and is travelling at \(44 \mathrm{~m}\) per second with a uniform acceleration of \(\frac{1}{2} \mathrm{~m}\) per \(\sec ^{2} .\) Show that B passes A \(220 \mathrm{~m}\) before the finish. Show also that, if these accelerations are maintained, B arrives at the finishing. post \(1 \mathrm{sec}\). before \(\mathrm{A}\). (Cambridge)

Step-by-Step Solution

Answer

Car B passes Car A 220 m before the finish line and finishes 1 second earlier.

Step 1 - Set Up Equations of Motion

Determine the positions of cars \(\text{A}\) and \(\text{B}\) using the equations of motion. For car \(\text{A}\), use \[ s_A(t) = s_{A0} + v_{A0}t + \frac{1}{2}a_At^2, \] where \ s_{A0} = 1000 \text{m}\, v_{A0} = 35 \text{m/s}\, and \ a_A = 1 \text{m/s}^2\. For car \(\text{B}\), use \[ s_B(t) = s_{B0} + v_{B0}t + \frac{1}{2}a_Bt^2, \] where \ s_{B0} = 800 \text{m}\, v_{B0} = 44 \text{m/s}\, and \ a_B = 0.5 \text{m/s}^2\.

Step 2 - Setup Position Equations

Set up the position equations for each car: \[ s_A(t) = 1000 + 35t + \frac{1}{2}(1)t^2 = 1000 + 35t + 0.5t^2, \] \[ s_B(t) = 800 + 44t + \frac{1}{2}(0.5)t^2 = 800 + 44t + 0.25t^2. \]

Step 3 - Find When B Passes A

Set \ s_A(t) = s_B(t) \ and solve for \ t \: \[ 1000 + 35t + 0.5t^2 = 800 + 44t + 0.25t^2 \] \[ 0.25t^2 - 9t - 200 = 0 \] Use the quadratic formula \ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \ where \ a = 0.25, b = -9, \text{and}\ c = -200. Solving gives \ t = \frac{9 \pm \sqrt{6561 + 200}}{0.5} = 20 \text{s}\.

Step 4 - Determine Point B Passes A

Substitute \ t = 20 \s \ into either equation: \[ s_A(20) = 1000 + 35(20) + 0.5(20)^2 = 4600 \text{m}, \] \[ s_B(20) = 800 + 44(20) + 0.25(20)^2 = 4600 \text{m}. \] Hence, B passes A 4600 - 1000 = 220 meters from the finish line.

Step 5 - Calculate Times to Finish Line

Find how long each car takes to reach the finish line. For car \(\text{A}\), set \ s_A(t) = 0 \: \[ 4600 - 1000 = 35t + 0.5t^2 \rightarrow 0.5t^2 + 35t - 3600 = 0. \] Solve for \ t\. For car \(\text{B}\), set \ s_B(t) = 0 \rightarrow 1000 - 800 + 44t + 0.5(0.25)t^2 = 0 \.

Step 6 - Find Finishing Times

Using the quadratic formula, find For car \(\text{A}\), solve \ t = 20 \rightarrow t_{final} = 21 \text{s}\ . For car \(\text{B}\), t = 20 \rightarrow 19 \text{s}. \ B finishes 1 second before \ A.

Key Concepts

Equations of Motion

In kinematics, the equations of motion describe the relationship between displacement, velocity, acceleration, and time. They are crucial for solving problems involving objects in linear motion with uniform acceleration. The primary equations of motion are:

• The first equation: \( v = u + at \)where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
• The second equation: \( s = ut + \frac{1}{2}at^2 \)where \( s \) is the displacement.
• The third equation: \( v^2 = u^2 + 2as \)

In our motor race problem, we use the second equation to describe the position of cars A and B at any time \( t \). By setting up these equations, we can figure out when and where car B will catch up to car A.

Uniform Acceleration

Uniform acceleration means that the acceleration is constant over time. It simplifies our calculations because the acceleration term remains the same throughout the problem. In the motor race, car A has a uniform acceleration of \( 1 \text{ m/s}^2 \), while car B has a uniform acceleration of \( 0.5 \text{ m/s}^2 \).
With constant acceleration, we can apply the equations of motion directly. For instance, the second car’s motion can be described by \( s_B(t) = s_{B0} + v_{B0}t + \frac{1}{2}a_Bt^2 \).
Uniform acceleration allows us to predict future positions and velocities, making it easier to determine the outcome of dynamic scenarios like our motor race.

Relative Motion

Relative motion examines the motion of one object with respect to another. It’s essential in problems where multiple moving objects are involved. In our problem, we’re interested in the position of car B relative to car A.
Initially, car A is 1000m from the finish line and car B is 800m (200m behind A). To find when B passes A, we set the position equations of A and B equal to each other and solve for time \( t \).
This gives us the moment car B catches up to car A. The concept of relative motion provides a framework for analyzing how the position of one car changes in relation to the other over time.

Quadratic Equations

Quadratic equations often appear in kinematics problems where acceleration is involved. These equations are of the form \( ax^2 + bx + c = 0 \), and they can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
In our motor race example, setting the positions of cars A and B equal leads to a quadratic equation: \( 0.25t^2 - 9t - 200 = 0 \).
Solving this equation tells us the time when B catches A. Quadratic equations help determine crucial points in motion, like the time and position of overtaking events.