Concept explainers
The Acela is an electric train on the Washington–New York–Boston run, carrying passengers at 170 mi/h. A velocity–time graph for the Acela is shown in Figure P2.46. (a) Describe the train’s motion in each successive time interval. (b) Find the train’s peak positive acceleration in the motion graphed. (c) Find the train’s displacement in miles between t = 0 and t = 200 s.
Figure P2.46 Velocity versus time graph for the Acela.
(a)
The description of the train’s motion in each successive time interval.
Answer to Problem 46P
Initially A is moving at constant positive velocity in the
Explanation of Solution
In the given velocity versus time graph of A, the curve is parallel to the time axis in the interval
From around
Conclusion:
Thus, initially A is moving at constant positive velocity in the
(b)
The train’s peak positive acceleration in the motion graphed.
Answer to Problem 46P
The train’s peak positive acceleration in the motion graphed is
Explanation of Solution
The slope of the graph in a given interval gives the acceleration of the object during it. The region of steepest slope in the
Write the equation for the acceleration of the object.
Here,
Write the equation for the slope.
Here,
Put the above equation in equation (I).
Conclusion:
Substitute
Therefore, the train’s peak positive acceleration in the motion graphed is
(c)
The train’s displacement in miles between
Answer to Problem 46P
The train’s displacement in miles between
Explanation of Solution
The cumulative area under the velocity versus time graph between
The velocity versus time graph is shown below.
Write the equation for the area of a rectangle.
Here,
Write the equation for the area of a triangle.
Here,
The area from
In figure 1, the length of the rectangle from
Substitute
Here,
The area from
In figure 1, the length of the rectangle from
Substitute
Here,
In figure 1, the base of the triangle from
Substitute
Here,
The area from
In figure 1, the length of the rectangle from
Substitute
Here,
In figure 1, the base of the triangle from
Substitute
Here,
Write the equation for the net displacement.
Here,
Conclusion:
Substitute
Therefore, the train’s displacement in miles between
Want to see more full solutions like this?
Chapter 2 Solutions
Principles of Physics: A Calculus-Based Text
- I am doing a free-fall experiment for Physics and I need to make a velocity vs. time graph using the average value of the final velocity. Do you know how I would be able to find the values for the velocity vs. time graph? The average value of the final velocity is 6.21954 m/s. It is question #9.arrow_forwardA KTM commuter train travels between two stations, A and B. The train starts from rest from station A and accelerates at a constant rate for 0.5 km until it reaches a speed of 30 m/s. It then travels at this constant speed for T s before decelerating at a constant rate of 1.5 m/s2 and coming to rest at station B. Figure 1 shows the velocity-time graph of the train’s journey between the two stations. Determine the value of T if the distance between the two stations 6 km. Calculate also the total time taken for the whole journey.arrow_forwardYou are driving home from school steadily at 99 km/h for 190 km . It then begins to rain and you slow to 63 km/h instantly. You arrive home after driving 5.0 hours. How far is your hometown from school?Express your answer using two significant figures. What was your average speed?Express your answer using two significant figures.arrow_forward
- A polar bear starts at the North Pole. It travels 1.0 km south, then 1.0 km east, and then 1.0 km north to return to its starting point. This trip takes 45 min. What was the bear's average speed and velocity?arrow_forwardThe acceleration of an object (in m/s2) is given by the function a ( t ) = 5 sin ( t ). The initial velocity of the object is v ( 0 ) = − 9 m/s. Round your answers to four decimal places. a) Find an equation v(t) for the object velocity. b) Find the object's displacement (in meters) from time 0 to time 3. c) Find the total distance traveled by the object from time 0 to time 3.arrow_forwardQ2. a) For the displacement equation, s = t2 +t-5, find the velocity and acceleration equations and plot each equation for the first 5 seconds. b) Compute your average velocity in the following two cases: (a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a speed of 3.05 m/s along a straight track. (b) You walk for 1.00 min at a speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a straight track. (c) Graph x versus I for both cases and indicate how the average velocity is found on the graph.arrow_forward
- The acceleration of an object increases linearly from 4 fps2to 12 fps2in 9 seconds. By the end of 9seconds, the velocity is 48 fps. NOTE: At the start, the object is at the zero-reference point ofdisplacement. a. Draw the a-t graph. Find the equation of the acceleration as the function of time.b. Draw the v-t graph. Find the equation of the velocity as the function of time.c. Draw the s-t graph. Find the equation of the displacement as the function of time.d. What is the initial velocity?e. What is the change in position during the 9 second interval?arrow_forwardA student performs a simple experiment to find the average acceleration of a falling object. He drops a baseball from a building and uses a string and meter stick to measure the height the ball was dropped. He uses a stopwatch to find an average time of fall for 3 trials from the same height and reports the following data: h =5.25 ± 0.15 m, t = 1.14 ± 0.06 S. a) Use the equation a = 2h/t2 to determine the average acceleration and its uncertainty. b) Comment on the accuracy of the acceleration result. Do you think the student made any mistakes? c) What one suggestion would you tell this student to improve the experimental result? Please explain.arrow_forwardAn object is at x = 0 at t = 0 and moves along the x axis according to the velocity-time graph in Figure P2.62. (a) What is the object’s acceleration between 0 and 4.0 s? (b) What is the object's acceleration between 4.0 s and 9.0 s? (c) What is the object's acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x = 0? (1) What is the final position x of the object at t = 18.0 s? (g) Through what total distance has the object moved between t = 0 and t = 18.0 s?arrow_forward
- A person standing on the roof of a building throws a ball directly upward. The ball misses the rooftop on its way down and eventually strikes the ground. The function s(t) = -16t2 + 64t + 80. describes the ball’s height above the ground, s(t), in feet, t seconds after it was thrown. Solve; a. Find the ball’s average velocity between the time it was thrown and 2 seconds later. b. Find the ball’s average velocity between 2 and 4 seconds after it was thrown. c. What do the signs in your answers to parts (a) and (b) mean in terms of the direction of the ball’s motion?arrow_forwardA car starts from rest and accelerates at a constant 10 m/s2 during aquarter-mile (402 m) race. How fast is the car going at the finish line? I know the answer to this problem is 90 m/s, but I don't know how the book i found it in came to this answer. How am I supposed to solve this problem without knowing either final velocity or time?arrow_forwardA student drives a moped along a straight road as described by the velocity–time graph in Figure P2.58. Sketch this graph in the middle of a sheet of graph paper. (a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (b) Sketch a graph of the acceleration versus time directly below the velocity–time graph, again aligning the time coordinates. On each graph, show the numerical values of x and ax for all points of inflection. (c) What is the acceleration at t = 6.00 s? (d) Find the position (relative to the starting point) at t = 6.00 s. (e) What is the moped’s final position at t = 9.00 s?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning