Finding Distance with Constant Acceleration

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Homework Help Overview

The problem involves a car accelerating at a constant rate and a truck moving at a constant speed. The goal is to determine the distance at which the car overtakes the truck after starting from rest.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the car's acceleration and the truck's constant speed, questioning the relevance of time in the context of the problem. There are attempts to clarify the distinction between matching speeds and covering equal distances.

Discussion Status

Some participants are exploring the correct interpretation of the problem, particularly regarding the significance of time and distance. There is an acknowledgment of confusion around the initial calculations and the need for clarity on what the problem is asking.

Contextual Notes

Participants note that the problem does not require determining the time it takes for the car to match the truck's speed, but rather when they are side-by-side, which may influence their approach to solving the problem.

B18
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Homework Statement


At the instant the traffic light turns green, a car starts with a constant acceleration of 5.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?



Homework Equations


Well I know the 4 kinematic equations for motion with constant acceleration.
I know final velocity of the truck=final velocity of the car.


The Attempt at a Solution


To find the time it took for the car to catch the truck i divided 70ft/s by 5.0ft/s^2=14.0s
t=14
Xi=0
Vi=0
a=5.0ft/s^2

Ive got the answers 490ft, 1015ft, 980 ft which are all incorrect. Am I using the wrong value for something here?

I appreciate any help.
 
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B18 said:

Homework Statement


At the instant the traffic light turns green, a car starts with a constant acceleration of 5.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?

Hi B18! The question DOES NOT ask how many seconds until the car's speed matches that of the truck!

The question asks: how many feet from their starting point are they seen to be side-by-side? Big difference
 
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NascentOxygen said:
Hi B18! The question DOES NOT ask how many seconds until the car's speed matches that of the truck!

The question asks: how many feet from their starting point are they seen to be side-by-side? Big difference

Hi NascentOxygen. Thanks for the reply. I am sorry I wasn't more specific. I found t=14 s. After plugging in the values I know I wasn't getting the correct answer.

What I mean is the t value I found correct? Did it take the car 14 s to reach the truck??

Wouldn't you need to know the duration of time in this problem??
 
I haven't worked it out, but I can see that dividing that v by that a will give you the time when the car reaches a speed of 70' per sec. The problem does not require you to determine anything relating to the event of the car's speed matching that of the truck. The problem involves equal distances, not equal speeds.
 

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