Finding the units in an algebraic equation for velocity

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SUMMARY

The discussion focuses on determining the units of coefficients a and b in the equation for velocity, v = at^3 + bt^4, where v is in meters per second (m/s) and t is in seconds (s). The correct units for a and b must ensure that the entire right-hand side of the equation also results in m/s. The conclusion is that a must have units of m/s^4 and b must have units of m/s^5 to satisfy the dimensional consistency required for the equation.

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Homework Statement


During a short interval of time the speed v in m/s of an automobile is given by v = at^3 + bt^4 , where the time t is in seconds. The units of a and b are respectively:
(a) m⋅s^2 ; m⋅s^4 (b) s^3/m; s^4/m (c) m/s^2 ; m/s^3 (d) m/s^3 ; m/s^4 (e) m/s^4 ; m/s^5

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The Attempt at a Solution


I have not a single clue, and I'm assuming that there is a trick to this that I'm not getting. I know that the velocity is measured in m/s, but I don't understand how finding the units of a and b is even possible.
 
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What do you know about the relationship between the units of terms in an equation where you are summing the terms? Can you add apples to oranges? :smile:
 
gneill said:
What do you know about the relationship between the units of terms in an equation where you are summing the terms? Can you add apples to oranges? :smile:
Nope, I know that you can only add apples to apples, or oranges to oranges, which means that both a and b need to be measured with the same units. Since this is an equation for velocity, I would assume that the units of the right hand side would also need to be m/s.
 
Kennedy said:
Nope, I know that you can only add apples to apples, or oranges to oranges, which means that both a and b need to be measured with the same units.
Think again. The terms also contain powers of t. What are the units associated with t?
Since this is an equation for velocity, I would assume that the units of the right hand side would also need to be m/s.
Right. So you need to choose the units of a and b so that their terms (including their t's) both work out to m/s.
 
gneill said:
Think again. The terms also contain powers of t. What are the units associated with t?

Right. So you need to choose the units of a and b so that their terms (including their t's) both work out to m/s.
So, I'm looking for units of a and b for which all of the s's in the numerator cancel, and there is only one s left in the denominator, and one m in the numerator?
 
Kennedy said:
So, I'm looking for units of a and b for which all of the s's in the numerator cancel, and there is only one s left in the denominator, and one m in the numerator?
Sounds like a plan!
 
gneill said:
Sounds like a plan!
Thank you!
 

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