Equation of uniformly accelerated motion

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SUMMARY

The discussion focuses on the equation of uniformly accelerated motion, specifically analyzing a ball thrown vertically upwards at an initial velocity of 16 m/s. The final velocity after 2 seconds is calculated using the formula v = u + at, resulting in -3.6 m/s, indicating downward motion. Displacement is debated, with one participant suggesting it is 0 due to the ball returning to its starting position, while another cites a displacement of 12.4 m using the equation r = ut + 1/2at². Clarifications on the signs of acceleration and units of measurement are also provided.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = u + at
  • Familiarity with the concept of acceleration due to gravity (9.8 m/s²)
  • Knowledge of displacement calculations using r = ut + 1/2at²
  • Basic grasp of units of measurement in physics (m/s, m/s²)
NEXT STEPS
  • Study the derivation and applications of kinematic equations in physics
  • Learn about the effects of gravity on projectile motion
  • Explore the concept of displacement versus distance in motion
  • Review common mistakes in unit conversions in physics
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of motion under uniform acceleration.

fcb
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Homework Statement



A ball is thrown vertically upwards at 16ms^-1. What is the velocity after 2 seconds
(acceleration is 9.8 m/s/s)

Homework Equations


v = u+at
V=velocity
u= Initial speed
a=acceleration down to earth
t=Time

The Attempt at a Solution



I don't quite understand it.
 
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v = u + at

v = final velocity, u = initial velocity, a = acceleration (i.e. gravity of 9.8 m/s^-1 downwards), t = time

v = 16 + (-9.8)(2)
= -3.6 m/s^-1, i.e. the ball is now traveling downwards.
 
dink87522 said:
v = u + at

v = final velocity, u = initial velocity, a = acceleration (i.e. gravity of 9.8 m/s^-1 downwards), t = time

v = 16 + (-9.8)(2)
= -3.6 m/s^-1, i.e. the ball is now traveling downwards.

Thanks so much

You're a life saver.
 
Ohh and is the displacement 51.6m?
 
I could be wrong on this, but wouldn't the displacement be 0? There is no motion in the horizontal plane, there is only motion in the vertical plane and the ball returns to its inital starting position, hence displacement is 0.
 
dink87522 said:
I could be wrong on this, but wouldn't the displacement be 0? There is no motion in the horizontal plane, there is only motion in the vertical plane and the ball returns to its inital starting position, hence displacement is 0.

Well the answer book said 12.4m

Using the equation: r=ut + 1/2at^2
r = displacement
u = Initial
t = time
a = acceleration
 
fcb said:
Ohh and is the displacement 51.6m?

Hi fcb! :smile:

You've got a sign wrong … this u and a have opposite signs. :wink:
 
tiny-tim said:
Hi fcb! :smile:

You've got a sign wrong … this u and a have opposite signs. :wink:

I thought to myself that can't be right because gravity is 9.8m/s/s relative. So maximum number in 2 seconds can only be 19.6m/s

thanks heaps lad.
 
@tiny-tim : Well his equation is kind of correct, as in fact it's when you substitute acceleration that the - sign comes up. Though you can easily forget to substitute the sign too.

@dink87522: You kinda confused acceleration units there, it's ms-2 instead of ms-1. I know it's a mistake though, don't worry.
 
  • #10
Is m/s/s the same as ms^-2?
 
  • #11
fcb said:
Is m/s/s the same as ms^-2?

(try using the X2 tag just above the Reply box :wink:)

Yes. :smile:
 
  • #12
I am on my phone so I can't use it. Well not I currently know of
 
  • #13
ah! you can type [noparse]"" and "" instead.[/noparse] :wink:
 

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