Calculating Distance Traveled in Varying Acceleration for Two Cars

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Homework Help Overview

The problem involves calculating the distance traveled by two cars, Car A and Car B, under varying acceleration and deceleration conditions. The scenario includes uniform acceleration and deceleration phases, as well as constant speed travel, all set along a straight road.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to determine the distance traveled by Car A during different phases of its journey, expressing confusion about the correct application of initial and final velocities. Participants discuss the meaning of variables in the equations and question the initial conditions used.

Discussion Status

Participants are actively engaging in clarifying the application of kinematic formulas and the definitions of variables. Some guidance has been offered regarding the correct use of initial velocity, but there is no explicit consensus on the approach for Car B, as one participant encourages the original poster to demonstrate further effort before receiving additional help.

Contextual Notes

There are forum rules in place that emphasize the importance of effort from the original poster before receiving direct answers. This may affect the flow of assistance provided in the discussion.

Gadget
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Hello everyone! I just registered to this board, and need some help solving a physics problem. It reads:

Car A leaves a city and travels along a straight road for 1.5 min at 60km/h. It then accelerates uniformly for 0.25 min until it reaches a speed of 80 km/h. It proceedes at that speed for 2.0 min, then decelerates uniformly for 0.50min until it comes to rest. Car B leaves the same city along the same road and accelerates uniformly for 1.6 min until it reaches a speed of 160km/h. It then decelerates uniformly until it comes to rest again after 1.6 min. How far will the two cars have traveled during the different stages?

This is as far as I completely understand it:

Car A:
1 step: conversion 60km/h = 1 km/min
therefore = 1.5 km

2 step: conversion 80km/h = 1.33 km/min
therefore v = vo + at
1.33 - 1 = a = 1.32km/min
--------
.25

here is confusion:
x = x0 + v0t + 1/2at^2

x = 1.5 + 1.33(.25 min) + (1/2)(1.32)(.25)^2
or
x = (1/2)(1.32)(.25)^2
?
 
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x = x0 + v0t + 1/2at^2

is the correct formula. But you got to know what the symbols mean. x_0 is the position of the object at the moment it starts accelerating. You chose x_0 well. Similarily, v_0 is the speed of the object, also at the moment it starts accelerating. For some reason you set v_0 = 1.33 km/min, which is the FINAL speed. Of course, you need v_0 = 1 km/min.
 
Thanks. What do I use for x_0 for car B?
 
I'm not going to tell you unless you show some effort. Sorry, forum rules.

https://www.physicsforums.com/showthread.php?t=28
 
I like this place, answers aren't just givin!
 

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