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
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
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
#a = x * t^p#
As you know, acceleration,
Velocity, on the other hand, tells you the rate at which the position of an object,
So, if position, or distance, has the dimension 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
#["L"] * ["T"]^(-2) = ["L"] * ["T"]^p#
At this point, it becomes clear that
#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)|)))#