# 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 **dimensionless quantity**, altogether to get

#a = x * t^p#

As you know, **acceleration**, **velocity** of an object,

**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 *dimensions* of

#v = ["L"] * ["T"]^(-1) -># distanceovertime

Consequently, acceleration will have *dimensions* of

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

#a = ["L"] * ["T"]^(-2) -># veloctiyovertime

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)|)))#