Vertical Motion: Understanding Velocity at a Given Time

[jdawg]

Homework Statement

An object is thrown vertically upward at 35 m/s. The velocity of the object 5.0 seconds later is:

Homework Equations

v_f=v_o-gt

The Attempt at a Solution

g=9.8 m/s
v_o=35 m/s
t=5s
v_f=?
v_f=35-(9.8)(5)=14 m/s down

I know this is the correct answer, but I'm confused about why you choose the equation above. The first time I worked this problem I tried to use v_f=v_o+gt. Could someone please explain to me why this would be wrong? Is it maybe because the acceleration is negative?

 

[BvU]

You're thinking the right thing. Point is: most of these equations are vector equations and vectors have a direction and a magnitude. By the time you need a number to compare with the answers in the back, you are dealing with magnitudes only.

Signs are established when you choose a coordinate system: if you say $\hat y$ (unit vector in positve y-direction) is upwards, then immediately the gravitation of the Earth $\vec g$ has a negative y-component -9.81 m/s², but he magnitude of g is still positive: 9.81 m/s² .

The velocity vector $\vec v$ in this coordinate system has a positive y-component ("object is thrown vertically upward ") and the acceleration "increases" the y-component with $\vec g_y$ m/s every second. Here $\vec g_y = -9.81$ m/s. For short (and in the beginning that can be genuinely confusing -- you are definitely not alone there --) we disregard the vector character and write scalar equations where we take the directions into account by using the proper signs. In this case you use v_f=v_o-gt which is fine if you understand it. It does mean that you end up with v_f < 0 . The silent understanding is that that means that v_f is the y-component of the upward velocity. After all you fill in a positive v₀, which still could be mistaken for a magnitude. But -14 is definitely not the magnitude of a vector.

35-(9.8)(5)=14 is not right if I ask my calculator. You correctly write "14 down", so you already do some interpreting there. Fine if you understand what you're doing, but dangerous in the beginning.

Tip: keep the units in there as long as you can:

35 m/s - 9.8 m/s² * 5 s = - 14 m/s can be checked wrt the numbers and wrt the dimensions. Your chances to discover and fix mistakes double !

 

[jdawg]

So would you just plug in the magnitudes into the equations? And the direction determines the sign used in the equation?

 

[nil1996]

Just use the concept that the up is positive and down is negative.Gravity acts down so it should have a negative sign.Similarly your ball is going upwards so the velocity of ball will have a positive sign.

 

[jdawg]

Ohhh ok, thanks guys!

 

[BvU]

:shy:$\Biggr\downarrow$