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Chapter 2: Problem 67

Two train cars are on a straight, horizontal track. One car starts at rest and is put in motion with a constant acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). This car moves toward a second car that is \(30.0 \mathrm{~m}\) away and moving at a constant speed of \(4.00 \mathrm{~m} / \mathrm{s}\). a) Where will the cars collide? b) How long will it take for the cars to collide?

Short Answer

Expert verified
Answer: a) The cars will collide at a position of \(25.0 \mathrm{~m}\) from Car 1's starting position. b) It will take \(5.0 \mathrm{~s}\) for the cars to collide.

Step by step solution

01

Equations of motion for Car 1

For Car 1, which starts from rest and accelerates at a constant rate of \(2.0 \mathrm{~m}/\mathrm{s}^2\), we can write the equation of motion as: \( x_1(t) = \frac{1}{2}at^2 \), where \(x_1(t)\) is the position of Car 1 as a function of time \(t\), \(a = 2.0 \mathrm{~m}/\mathrm{s}^2\) is the acceleration, and \(t\) is time.
02

Equations of motion for Car 2

For Car 2, moving at a constant speed of \(4.0 \mathrm{~m}/\mathrm{s}\) and initially \(30.0 \mathrm{~m}\) away from Car 1, we can write the equation of motion as: \( x_2(t) = 30.0 \mathrm{~m} + vt \), where \(x_2(t)\) is the position of Car 2 as a function of time \(t\), \(v = 4.00 \mathrm{~m}/\mathrm{s}\) is the constant speed, and \(t\) is time.
03

Find the collision point

The cars will collide when their positions are equal, so we can set the two position functions equal to each other: \( \frac{1}{2}at^2 = 30.0 \mathrm{~m} + vt \) Plugging the acceleration of Car 1 and the speed of Car 2 into the equation, we get: \( \frac{1}{2}(2.0 \mathrm{~m}/\mathrm{s}^2)t^2 = 30.0 \mathrm{~m} + (4.00 \mathrm{~m}/\mathrm{s})t \)
04

Solve for the time it takes for the cars to collide

To solve for \(t\), we can simplify the given equation: \( t^2 = 30.0 \mathrm{~m} + 4.00 \mathrm{~m}/\mathrm{s}t \) Now, combine the terms involving \(t\) and factor out \(t\) from the resulting expression to find the time: \( t^2 - 4.00\mathrm{~m}/\mathrm{s}t = 30.0 \mathrm{~m} \) \( t(t - 4.00\mathrm{~m}/\mathrm{s}) = 30.0 \mathrm{~m} \) Because \(t\geq0\), we get: \( t = 4.00\mathrm{~m}/\mathrm{s} - 30.0 \mathrm{~m} = 5.0 s \)
05

Calculate the collision point

Now that we've found the time it takes for the cars to collide, we can use either position equation to find the position of the collision. We'll use the equation for Car 1: \( x_1(t) = \frac{1}{2}(2.0 \mathrm{~m}/\mathrm{s}^2)(5.0 \mathrm{~s})^2 \) \( x_1(5.0 \mathrm{~s}) = 25.0 \mathrm{~m} \) Therefore, the cars will collide at a distance of \(25.0 \mathrm{~m}\) from Car 1's starting position. a) The cars will collide at a position of \(25.0 \mathrm{~m}\) from Car 1's starting position. b) It will take \(5.0 \mathrm{~s}\) for the cars to collide.

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