A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the speedup and slowdown stages respectively? O 6.71 m/s^2, then 3.43 m/s^2 5.26 m/s^2, then 10.29 m/s^2

College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter2: Motion In One Dimension
Section: Chapter Questions
Problem 4CQ: (a) Can the equations in Table 2.4 be used in a situation where the acceleration varies with time?...
icon
Related questions
icon
Concept explainers
Topic Video
Question

A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the speedup and slowdown stages respectively?

A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate
over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the
speedup and slowdown stages respectively?
O 6.71 m/s^2, then 3.43 m/s^2
5.26 m/s^2, then 10.29 m/s^2
2.27 m/s^2, then 2.27 m/s^2
10.29 m/s^2, then 5.26 m/s^2
Transcribed Image Text:A car starts from rest, then accelerates at a constant rate over a distance of 72 m. It then immediately decelerates at a constant rate over a distance of 141 m. The entire trip lasts a total duration of 13.7 s. What were the magnitudes of the car s accelerations for the speedup and slowdown stages respectively? O 6.71 m/s^2, then 3.43 m/s^2 5.26 m/s^2, then 10.29 m/s^2 2.27 m/s^2, then 2.27 m/s^2 10.29 m/s^2, then 5.26 m/s^2
Expert Solution
Step 1

Given 

Initial velocity u = 0 

Distance travelled during acceleration = 72 m 

Distance covered during deacceleraration = 141 m 

Total time taken = 13.7 s  

Area under velocity time graph represents displacement. 

trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Knowledge Booster
Displacement, velocity and acceleration
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781285737027
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning