Acceleration is related to distance and time by the following expression: #a = 2xt^p#. Find the power #p# that makes this equation dimensionally consistent?

1 Answer
Sep 2, 2016

Answer:

#p=-2#

Explanation:

All you have to do here is replace the quantities given to you in that equation with their corresponding dimensions, which as you know are

https://physicsforus.wordpress.com/physics-1/magnitude-and-units/c-dimensi/

So, you know that you have

  • #"distance" = x -> ["L"]#
  • #"time" = t -> ["T"]#

Now, your equation looks like this

#a = 2 * x * t^p#

Since you must match dimensions here, you can eliminate the #2#, a dimensionless quantity, altogether to get

#a = x * t^p#

As you know, acceleration, #a#, tells you the rate at which the velocity of an object, #v#, changes with respect to time, #t#.

Velocity, on the other hand, tells you the rate at which the position of an object, #x#, changes with respect to a given frame of reference, and also with respect to time.

So, if position, or distance, has the dimension of #["L"]# and time has the dimension of #["T"]#, you can say that velocity will have the dimensions of

#v = ["L"] * ["T"]^(-1) -># distance over time

Consequently, acceleration will have dimensions of

#a = ["L"] * ["T"]^(-1) * ["T"]^(-1)#

#a = ["L"] * ["T"]^(-2) -># veloctiy over time

This means that the left side of the equation is

#["L"] * ["T"]^(-2) = x * t^p#

On the right side of the equation, replace #x# and #t# with their respective dimensions to get

#["L"] * ["T"]^(-2) = ["L"] * ["T"]^p#

At this point, it becomes clear that #p=-2#, since

#color(red)(cancel(color(black)(["L"]))) * ["T"]^(-2) = color(red)(cancel(color(black)(["L"]))) * ["T"]^p#

#["T"]^(-2) = ["T"]^p implies color(green)(bar(ul(|color(white)(a/a)color(black)(p = -2)color(white)(a/a)|)))#