Auto & Truck Movements: Solving for Overtaking Distance & Speed

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Homework Help Overview

The problem involves an automobile accelerating from rest and a truck moving at a constant speed, with the goal of determining the distance at which the automobile overtakes the truck and its speed at that moment. The subject area includes kinematics and motion analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to describe the motion of both vehicles, with one focusing on accelerated motion and the other on constant speed. There is a mention of deriving formulas and questioning the reasoning behind them.

Discussion Status

Some participants have offered insights into the derivation of kinematic equations and how they relate to the problem. There is an acknowledgment of the connections between the equations used, but no consensus on the methods or outcomes has been reached.

Contextual Notes

One participant references answers found in the back of a book, which may influence their approach to the problem. There is a discussion about the dimensions involved in the equations, indicating a focus on understanding the underlying principles rather than simply applying formulas.

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"At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s^2. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile. (a). How far beyond the traffic signal will the automobile overtake the truck? (b). How fast will the automobile be traveling at that instant?"
 
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Start by writing the kinematic formulas describing the distance of each as a function of time. One formula will involve accelerated motion; the other, constant speed motion.
 
hmm... I really think I have lost brain cells or something. In the back of the book it has the answers (a). 82m (b). 19 m/s

I get 82m if I use x=(2v^2)/a. But, I'm not sure how to derive that formula.

For the (b), I used v^2 = Vinitial^2 +2a(X-Xinitial) and got 19 m/s
 
The first equation is derived from the one you used in b.
 
whozum said:
The first equation is derived from the one you used in b.

oh wow, you're right. I never thought that a times x was just v^2. I was not thinking of the dimensions.

So after moving v initial over to the other side, I had delta V^2 = 2 v^2

Then i broke up delta V^2 into v times v. I change one of those into x/1 and 1/t. I combined v and 1/t to get a. So now I had a times x = 2v^2.

I get the first equation of my previous post after dividing by a on both sides.

x = (2v^2)/a


thanks, to both of you
 

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