Quick question regarding speed & constant acceleration

  • #1
NutriGrainKiller
61
0
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
 

Answers and Replies

  • #2
courtrigrad
1,236
1
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:
  • #3
NutriGrainKiller
61
0
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
 
  • #4
courtrigrad
1,236
1
yes you assume starting position is 0.
 

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