Two cars drive on a straight highway. At time t = 0, car 1 passes mile marker 0 traveling due east with a speed of 22.0 m/s. At the same time, car 2 is 1.4 km east of mile marker 0 traveling at 30.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 3.0 m/s2 and car 2 is slowing down with an acceleration of magnitude 2.5 m/s2.
could you do a step by step process on how to solve the problem. you don't have to give the answer but how to arrive to the answer would be great
What is your question about this problem? I will be happy to critique your thinking. We are not in the habit of doing homework or tests for students.
To solve this problem, we need to understand the motion equations of the two cars and analyze their movements.
First, let's calculate the time it takes for car 1 to reach car 2. We can use the formula:
\[ t = \frac{d}{v} \]
where \( t \) is time, \( d \) is distance, and \( v \) is velocity. The distance between car 1 and car 2 is given as 1.4 km or 1400 m. Car 1 has a velocity of 22.0 m/s, so the time taken for car 1 to reach car 2 is:
\[ t_{12} = \frac{1400 \, \text{m}}{22.0 \, \text{m/s}} \]
Simplifying the equation, we get:
\[ t_{12} \approx 63.64 \, \text{s} \]
So, it takes approximately 63.64 seconds for car 1 to reach car 2.
Now, let's calculate the final velocity of car 1 when it reaches car 2. We can use the equation:
\[ v_f = v_i + at \]
where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. We know that the initial velocity of car 1 is 22.0 m/s, the acceleration is 3.0 m/s², and the time is 63.64 s. Substituting the values into the equation:
\[ v_f = 22.0 \, \text{m/s} + (3.0 \, \text{m/s²})(63.64 \, \text{s}) \]
Simplifying the equation, we get:
\[ v_f \approx 209.92 \, \text{m/s} \]
So, when car 1 reaches car 2, it has a final velocity of approximately 209.92 m/s.
Next, let's calculate the distance traveled by car 2 when it is overtaken by car 1. We can use the equation:
\[ d = v_i t + \frac{1}{2} a t^2 \]
where \( d \) is the distance traveled, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken. We know that the initial velocity of car 2 is 30.0 m/s, the acceleration is -2.5 m/s² (negative because it's deceleration), and the time is 63.64 s. Substituting the values into the equation:
\[ d = (30.0 \, \text{m/s})(63.64 \, \text{s}) + \frac{1}{2} (-2.5 \, \text{m/s²})(63.64 \, \text{s})^2 \]
Simplifying the equation, we get:
\[ d \approx 1805.46 \, \text{m} \]
So, car 2 travels approximately 1805.46 meters before it is overtaken by car 1.
To summarize:
- Car 1 takes approximately 63.64 seconds to reach car 2.
- When car 1 reaches car 2, its final velocity is approximately 209.92 m/s.
- Car 2 travels approximately 1805.46 meters before it is overtaken by car 1.