Check work on kinematics problem

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Homework Help Overview

The problem involves kinematics, specifically the motion of a truck that accelerates from rest to a specified speed, maintains that speed, and then accelerates again. The original poster seeks to verify their calculations regarding the time taken to reach a final speed and the distance traveled during this time.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster presents their calculations for time and distance, expressing discrepancies with the book's answers. Some participants request to see the original poster's work to identify potential errors.

Discussion Status

The discussion is ongoing, with participants engaging in a review of the original poster's calculations. There is an acknowledgment of the original poster's concerns about the reliability of their textbook, and some guidance is being offered to clarify the calculations.

Contextual Notes

The original poster mentions that the textbook has had numerous mistakes, which may influence their confidence in the provided answers. There is also a humorous remark about the acceleration rate of the truck, suggesting a questioning of the problem's parameters.

caperjay
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Homework Statement


A truck accelerates uniformly from rest to 48km/h in 20s, holds speed for 2 minutes, and then accelerates to 100km/h at a rate of 17m/s^2. How long does it take, from rest, to reach 100km/h and what distance it traveled in this time?


The book says it takes 148.5s, and I keep getting 140.85s.
For distance the book says 1904m, and I been getting lower at 950.57m.

I was hoping someone could just check the answers for me, since I am doing a course from a book that has had a lot of mistakes thus far, if it turns out that my answers are wrong I will post my what I've been doing to get some help on it. Thanks much!
 
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Hey Caperjay, could you post some work? Then we'll see what's up.
 
mateomy said:
Hey Caperjay, could you post some work? Then we'll see what's up.

Heres what I've done. For the total time I got 20s + 120s + (13.33=27.78 + 17m/s^2xt) where t turns out to be 0.85s, for a total of 140.85s.

For total distance I've got (0+13.33)/2 x 20s + (13.33 + 13.33)/2 x 120s + (13.33 + 27.78)/2 x 0.85s, for a total of 950.57m.
 
"at a rate of 17m/s^2"

Mighty powerful truck!
 

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