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Chapter 2: Problem 33

The St. Charles streetcar in New Orleans starts from rest and has a constant acceleration of \(1.20 \mathrm{m} / \mathrm{s}^{2}\) for \(12.0 \mathrm{s} .\) (a) Draw a graph of \(v_{x}\) versus \(t .\) (b) How far has the train traveled at the end of the \(12.0 \mathrm{s} ?\) (c) What is the speed of the train at the end of the \(12.0 \mathrm{s} ?\) (d) Draw a motion diagram, showing the streetcar's position at \(2.0-\mathrm{s}\) intervals.

Short Answer

Expert verified
Answer: The speed of the streetcar at the end of the 12.0 seconds is 14.4 m/s. The total distance it traveled during that time is 86.4 meters.

Step by step solution

01

a) Drawing a graph of vx versus t

Since the streetcar starts from rest and has a constant acceleration, we will have a straight line starting from the origin (0,0) with a positive slope equal to the acceleration: 1.20 m/s虏. The graph should look like an inclined straight line, representing a linear function.
02

b) Distance traveled at the end of the 12.0 s

We will use the second kinematic equation to find the final position: \(x = x_0 + v_0t + \frac{1}{2}at^2\). The initial position x0 is 0, and the initial velocity v0 is also 0. The equation reduces to \(x = \frac{1}{2}at^2\). Plugging in the values, we get: \(x = \frac{1}{2}(1.20 m/s虏)(12.0s)^2 = 3.60 m * 144 = 86.4 m\). So, the streetcar traveled 86.4 meters at the end of the 12.0 s.
03

c) Speed at the end of the 12.0 s

We use the first kinematic equation to find the final velocity: \(v_x = v_0 + at\). The initial velocity v0 is 0. Plugging in the values, we get: \(v_x = (1.20 m/s虏)(12.0s) = 14.4 m/s\). So, the speed of the train at the end of the 12.0 s is 14.4 m/s.
04

d) Drawing a motion diagram at 2.0-s intervals

We will use \(x = \frac{1}{2}at^2\) to find the positions at 2.0-s intervals. Using the time values 2, 4, 6, 8, 10, and 12, we obtain the positions 2.4 m, 9.6 m, 21.6 m, 38.4 m, 60.0 m, and 86.4 m respectively. In a motion diagram, start with a dot at the initial position and label it as 0 s. Then draw a dot for each of the positions and label them accordingly: 2 s, 4 s, 6 s, 8 s, 10 s, and 12 s. Note that the distance between the dots on the motion diagram increases over time since the streetcar is accelerating.

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Most popular questions from this chapter

A car is speeding up and has an instantaneous velocity of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction when a stopwatch reads \)10.0 \mathrm{s} .$ It has a constant acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) -direction. (a) What change in speed occurs between \(t=10.0 \mathrm{s}\) and \(t=12.0 \mathrm{s} ?(\mathrm{b})\) What is the speed when the stopwatch reads \(12.0 \mathrm{s} ?\)
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