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Chapter 2: Problem 43

A train, traveling at a constant speed of \(22 \mathrm{m} / \mathrm{s}\), comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude $1.4 \mathrm{m} / \mathrm{s}^{2} .\( (a) Draw a graph of \)v_{x}\( versus \)t\( where the \)x$ -axis points up the incline. (b) What is the speed of the train after $8.0 \mathrm{s}$ on the incline? (c) How far has the train traveled up the incline after \(8.0 \mathrm{s} ?\) (d) Draw a motion diagram, showing the trains position at 2.0 -s intervals.

Short Answer

Expert verified
Answer: The train's speed after 8 seconds on the incline is 10.8 m/s, and it has traveled a distance of 131.2 meters during that time.

Step by step solution

01

Drawing the Velocity-Time Graph

To draw the velocity-time graph, we need to consider the given initial speed of the train and its constant acceleration up the incline. On the x-axis (time in seconds), plot a point at (0, 22), representing the initial speed of the train. Since the train is slowing down, the acceleration is negative, so we know the graph will have a negative slope. Calculate the slope by dividing the acceleration by time: \(-1.4/1 = -1.4\). Draw a straight line with a slope of -1.4 ms^-2, starting from the initial point. This line represents the velocity of the train over time.
02

Calculate the Speed After 8 seconds

Use the formula for final velocity with constant acceleration: \(v_f = v_0 + at\) Where \(v_f\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time interval given (8 seconds). Plugging the values: \(v_f = 22 - 1.4(8)\) \(v_f = 22 - 11.2\) \(v_f = 10.8 \mathrm{m}/\mathrm{s}\) So, the speed of the train after 8 seconds on the incline is \(10.8 \mathrm{m}/\mathrm{s}\).
03

Determine the Distance Traveled Up the Incline After 8 Seconds

Use the formula for position with constant acceleration: \(x = x_0 + v_0t + \frac{1}{2}at^2\) Where \(x\) is the position after time \(t\), \(x_0\) is the initial position (0 in this case), \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time interval (8 seconds). Plugging the values: \(x = 0 + 22(8) + \frac{1}{2}(-1.4)(8)^2\) \(x = 176 - 44.8\) \(x = 131.2\) meters Thus, the train has traveled \(131.2 \mathrm{m}\) up the incline after 8 seconds.
04

Drawing the Motion Diagram

To draw the motion diagram, consider the train's position at 2-second intervals. Use the position formula mentioned in step 3 to find the position of the train at these time intervals. To create the motion diagram, plot these positions on the graph, marking the positions at 0, 2, 4, 6, and 8 seconds intervals. The distances between points will become shorter due to the train's acceleration, indicating that the train's speed decreases as it goes up the incline. Connect the points with a smooth curve representing the train's motion during the 8 seconds on the incline. This motion diagram shows the positions of the train at 2-second intervals throughout its trajectory.

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