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Homework Help: Acceleration Conditions

  1. Nov 5, 2014 #1
    • Warning: Template must be used for homework help.
    An object has initial speed u and acceleration a. After travelling a distance s, its final speed is v.
    Which of the following includes the two conditions necessary for the equation, v^2 = u^2 +2as, to apply?

    I know the answer is a has constant magnitude and a has constant direction, but I am not sure why. Could someone please explain this to me?
    Last edited: Nov 5, 2014
  2. jcsd
  3. Nov 5, 2014 #2


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    Hello air-in,

    Welcome to PF! :)

    Before we proceed further, what level of mathematics is used in your class/coursework? Is calculus required for your coursework material? How about differential equations?

    Is the "why" part of the assigned problem, or are you just curious?
  4. Nov 5, 2014 #3
    I am currently taking calculus, and we have used differential equations. However, I believe either Algebra 2/Trig or Precalc is required for the course.
    I am choosing to do test corrections, and the "why" part is mandatory.
  5. Nov 5, 2014 #4


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    This might get a little difficult without calculus, since the equation was derived using calculus, under the assumption that the acceleration is uniform. Never fear though, it may be possible to explain without calculus.

    You can derive the uniform acceleration, kinematics equation, [itex] v^2 = u^2 + 2as [/itex] by combining other kinematics equations for uniform acceleration. I'll let you do that yourself, but first I'd like to take a look at one of the more simple, uniform acceleration, kinematics equations.

    Consider another uniform acceleration, kinematics equation:

    [tex] v = u + at [/tex]

    Make a graph of [itex] v [/itex] vs. [itex] t [/itex]. Notice that it's a straight line (of the form y = mx + b, with simple substitutions).

    What is the slope of the line? How does the slope of that line relate to acceleration?

    (And what is the area under the curve? [we might come back to this later])

    Can you see how that [itex] v = u + at [/itex] would not necessarily hold true if the value of the acceleration varied with time? [Assuming you only get to use a single value for a].
    Last edited: Nov 5, 2014
  6. Nov 5, 2014 #5


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    In my last post I made mention of calculating the area under the curve. It turns out that is pretty important to the "why" part.

    Recall that [itex] v = u + at [/itex] is a straight line. For the this exercise of finding the area, assume that both [itex] u [/itex] and [itex] a [/itex] are positive.

    So the area under that line is a triangle on top of a rectangle.

    The rectangle has a length of [itex] t [/itex] and a height of [itex] u [/itex].

    The triangle on top of it has length of [itex] t [/itex] and a height of [itex] at [/itex].

    So what's the total area? (i.e. what is the sum of the areas of the triangle and rectangle)?

    This process of finding the area under the line might seem like a lot of busywork, but it is important for the next step when we come back to the [itex] v^2 = u^2 +2as [/itex] equation.

    What assumption did you make when calculating the area under the line? Would the answer be the same if the line was not straight?
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