Vertical straight line motion

  • Thread starter jdawg
  • Start date
  • #1
367
2

Homework Statement



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

Homework Equations



vf=vo-gt

The Attempt at a Solution


g=9.8 m/s
vo=35 m/s
t=5s
vf=?
vf=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 vf=vo+gt. Could someone please explain to me why this would be wrong? Is it maybe because the acceleration is negative?
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
14,400
3,712
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 vf=vo-gt which is fine if you understand it. It does mean that you end up with vf < 0 . The silent understanding is that that means that vf 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 !
 
  • #3
367
2
So would you just plug in the magnitudes into the equations? And the direction determines the sign used in the equation?
 
  • #4
301
7
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.
 
  • #5
367
2
Ohhh ok, thanks guys!
 
  • #6
BvU
Science Advisor
Homework Helper
14,400
3,712
:shy:##\Biggr\downarrow##
 

Related Threads on Vertical straight line motion

  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
9
Views
1K
Top