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

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 #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)|)))#