Quick question regarding speed & constant acceleration

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Homework Help Overview

The discussion revolves around a physics problem involving an antelope moving with constant acceleration, covering a distance of 75.0 m in 7.60 s, with a final speed of 14.0 m/s. Participants are exploring the kinematic equations relevant to finding the initial speed and acceleration of the antelope.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the problem by identifying the known variables and expressing the relationships using kinematic equations. Some participants suggest using specific equations to relate distance, time, and speed, while others question the assumption of the starting position being zero.

Discussion Status

Participants are actively engaging with the problem, discussing the application of kinematic equations and confirming assumptions about initial conditions. There is a focus on clarifying the setup of the problem without reaching a definitive conclusion.

Contextual Notes

There is mention of the original poster's learning challenges, which may influence the pace and depth of the discussion. The problem is framed within the context of a university-level physics course, indicating a structured approach to problem-solving.

NutriGrainKiller
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I'm doing homework from the first week of Physics 111 in the 2nd chapter (this is university calc-based physics by the way). I am a bit of a slow learner (dyslexic :frown: ), so take it easy on me. I know the answer to the first question, but I would like to understand it. Here is the problem:

An antelope moving with constant acceleration covers the distance 75.0 m between two points in time 7.60 s. Its speed as it passes the second point is 14.0 m/s.

Q1): What is its speed at the first point?
A1): 5.74 m/s

Q2): What is the acceleration?

Thanks guys
 
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Ok so for this type of problem you have to use your kinematic equations. We are given a distance between two points, the time it takes for the antelope to cover that distance, and the speed when it passes the second point. So we need to use an equation that contains all these variables. We would use [itex]x = x_{0} + \frac{1}{2}(v_{x}_{0}+v_{x})t[/itex]. We could rewrite this as [itex]x - x_{0} = \frac{1}{2}(v_{x}_{0}+v_{x})t[/itex]. [itex]x - x_{0}[/itex] is really the distance, so [itex]75 = \frac{1}{2}(v_{x}_{0} + 14)(7.60)[/itex]. So just solve for [itex]v_{x}[/itex]. For the second question, use the equation [itex]x - x_{0} = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}[/itex]. [itex]x_{0} = 0[/itex] and you know [itex]v_{x}_{0}[/itex] and [itex]t[/itex] from the previous question. So just solve for [itex]a[/itex].
 
Last edited:
courtrigrad said:
Ok so for this type of problem you have to use your kinematic equations. We are given a distance between two points, the time it takes for the antelope to cover that distance, and the speed when it passes the second point. So we need to use an equation that contains all these variables. We would use [itex]x = x_{0} + \frac{1}{2}(v_{x}_{0}+v_{x})t[/itex]. We could rewrite this as [itex]x - x_{0} = \frac{1}{2}(v_{x}_{0}+v_{x})t[/itex]. [itex]x - x_{0}[/itex] is really the distance, so [itex]75 = \frac{1}{2}(v_{x}_{0} + 14)(7.60)[/itex]. So just solve for [itex]v_{x}[/itex]. For the second question, use the equation [itex]x - x_{0} = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}[/itex]. [itex]x_{0} = 0[/itex] and you know [itex]v_{x}_{0}[/itex] and [itex]t[/itex] from the previous question. So just solve for [itex]a[/itex].

do i just assume that X0 (subscript 0) is equal to 0? Like I assume the starting position is 0 in other words
 
yes you assume starting position is 0.
 

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