Misleading Textbook Equation for vf^2=vi^2 + 2ad

[PhysTeacher88]

The textbook (Nelson 11) at my school lists the "big 5" equations for uniform acceleration. In all but one, they use vectors.

For vf^2=vi^2 + 2ad, the opt not to use vectors.

Is there a deep reason why we would not want to use the vectors?

I understand that when you square the velocity, the direction information is lost, however, without making the acceleration and the displacement vectors (textbook reads distance because it's not a vector), students will not get questions like this correct:

A ball is thrown up at 10m/s, how high will it go?

If I treat everything as scalar, we get:

(0m/s) = (10m/s)^2 + (9.8)d

the distance ends up being a negative value, which is clearly not true given the context.

If they at least made the "a" and delta "d" vectors, they would not run into this problem.

I'm guess they took this formula from conservation of energy, rather than thinking about this from a kinematics perspective.

Am I missing something, or is this an oversight?

Cheers,

K

 

[Mark44]

The textbook (Nelson 11) at my school lists the "big 5" equations for uniform acceleration. In all but one, they use vectors.

For vf^2=vi^2 + 2ad, the opt not to use vectors.

Is there a deep reason why we would not want to use the vectors?

I understand that when you square the velocity, the direction information is lost, however, without making the acceleration and the displacement vectors (textbook reads distance because it's not a vector), students will not get questions like this correct:

A ball is thrown up at 10m/s, how high will it go?

If I treat everything as scalar, we get:

(0m/s) = (10m/s)^2 + (9.8)d

the distance ends up being a negative value, which is clearly not true given the context.

If they at least made the "a" and delta "d" vectors, they would not run into this problem.

I'm guess they took this formula from conservation of energy, rather than thinking about this from a kinematics perspective.

Am I missing something, or is this an oversight?

Cheers,

K

What you're missing is that you have the initial velocity, $v_i$, being positive, which means that $a$ must be negative. So here $a = -9.8 m/sec^2$.

Also, the probable reason for not using vectors is that they are dealing with motion in one dimension.

 

[Frabjous]

It’s a special case of the work-energy theorem which is a scalar equation.
ΔKE= net Work (=F⋅d for constant F)

 

[andresB]

What you're missing is that you have the initial velocity, $v_i$, being positive, which means that $a$ must be negative. So here $a = -9.8 m/sec^2$.

Correct.

Also, notice that d also have a sign. It is positive if the motion was upward, and negative if the motion was downward.