Can Mixed Units in Ratios Be Converted and Used Accurately in Calculations?

[photon184739]

Homework Statement

The driver of a car moving at 90.0 km/h presses down on the brake as the car enters a circular curve of radius 150.0 m. If the speed of the car is decreasing at a rate of 9.0 km/h each second, what is the magnitude of the acceleration of the car at the instant its speed is 60.0 km/h?

(Openstax Physics 1 Chapter 4 q. 81)

Relevant Equations

a_c = v^2/r (centripetal acceleration)
a = a_c + a_t (centripetal and tangential acceleration)

I have the solution, but my question is that in my first attempt, I simply calculated a_c = 60^2/150 ignoring the fact that the units for 60 are km/hr and the units for 150 are meters. Is this ever salvageable? Is there any way to retain the 'mixed' units of km with meters in the ratio and apply a conversion factor, or is information lost and the approach should be 'scrapped'?

 

[hmmm27]

Why don't you write the first part out in long form, including all units. See where that takes you. Then work on combining the two accelerations.

 

[kuruman]

a = a_c + a_t (centripetal and tangential acceleration)

Also, this equation is incorrect. Quantities a_c and a_t are orthogonal components of the acceleration vector. Simply adding them is incorrect if your goal is to find the magnitude.

 

[Lnewqban]

You need to do something about the hour also, unless you want your acceleration to be $m/h^2$.
The best way is to convert kilometers to meters and hours to seconds before entering values into the equation.

 

[etotheipi]

As has been mentioned, if in doubt, carry the units through your calculation. $$\frac{(60 \text{km}\text{h}^{-1})^2}{150\text{m}} = \frac{3600 \times (10^3 \text{m})^2 \times (3600s)^{-2}}{150 \text{m}} = \text{etc.}$$ @Lnewqban's method has the benefit that if you always use SI units, the units will simplify into the standard SI unit for the new calculated quantity, which can save lot of work!

 

[kuruman]

As has been mentioned, if in doubt, carry the units through your calculation. $$\frac{(60 \text{km}\text{h}^{-1})^2}{150\text{m}} = \frac{3600 \times (10^3 \text{m})^2 \times (3600s)^{-2}}{150 \text{m}} = \text{etc.}$$ @Lnewqban's method has the benefit that if you always use SI units, the units will simplify into the standard SI unit for the new calculated quantity and that means we don't need to worry much about the units except for giving the final answer.

Perhaps. Since the question has already chosen the units by giving the tangential acceleration not in SI but as $a_t=9.0~\mathrm{\dfrac{km/hr}{s}}$, I would say that the magnitude of the acceleration should be given in the same units. All one has to do is convert the centripetal acceleration as calculated by OP, $a_c=\dfrac{(60~\mathrm{km/hr)^2}}{150~\mathrm m}$ into these same units.

 

[etotheipi]

Perhaps. Since the question has already chosen the units by giving the tangential acceleration not in SI but as $a_t=9.0~\mathrm{\dfrac{km/hr}{s}}$, I would say that the magnitude of the acceleration should be given in the same units. All one has to do is convert the centripetal acceleration as calculated by OP, $a_c=\dfrac{(60~\mathrm{km/hr)^2}}{150~\mathrm m}$ into these same units.

Whoops, I didn't even see that part of the question. When we do the addition in $a = \sqrt{a_t^2 + a_n^2}$ the two evidently have to have the same units in order to 'factorise the units out' in the addition.