Kinematic Equations: Position & Velocity at t=2.40s

In summary, using the given kinematic equations and initial values, it is possible to determine the position and velocity at t=2.40 s. The horizontal velocity of the initial velocity can be found by multiplying the initial velocity (25.0 m/s) by the cosine of the given angle (65.0 degrees). To find the horizontal/vertical velocity at 2.40 s, the vector can be broken down into its components and then calculated using appropriate formulas. Finally, the horizontal position at 2.40 s can be found by multiplying the initial velocity (25.0 m/s) by the given time (2.40 seconds). To find the velocity at 2.40 s, the formula V = V0 +
  • #1
Throwback24
15
0

Homework Statement



Using the appropriate kinematic equations and initial values, determine the position and velocity at t=2.40 s.

(x, y)=0, 0
(vi, 0)=(25.0 m/s, 65.0 degree)

a.Horizontal Velocity of the initial velocity

b.Horizontal/Vertical Velocity at 2.40 s

c.Horizontal position at 2.40 s

Homework Equations





The Attempt at a Solution



a.(25 m/s)(Cos 65)

b.

c. X=(25 m/s)(2.40 seconds)

How do I get Horizontal/Vertical Velocity at 2.40 seconds?
 
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  • #2
someone correct me if i am wrong, but i believe that you would want to break the horizontal/vertical velocity vector into into its components and then use either trig or pythag to find the avg velocity.
 
  • #3
But it's asking for the velocity @ 2.40 seconds. Don't you break up vectors into components?
 
  • #4
For something moving 25ms-1 at an angle of 65 degrees to the horizontal you use. 25cos65 for the horizontal and 25sin65 for the vertical. use appropriate formula for the given times to find its velocity
 
  • #5
I'm wrong on C, it has to be:

25 m/s (Cos 65)(2.40 m)

I still don't get B. Gregg, isn't that how I solve for Initial velocity?
 
  • #6
You can use
[tex]v = u + at[/tex] you know u, you know a and you are given t.
 
  • #7
I know V is Velocity, A is Acceleration and T is time. What is U? I've never seen it.
 
  • #8
u is initial velocity, v0
 
  • #9
Vf = Vi + at
or
V = V0 + at
 

Related to Kinematic Equations: Position & Velocity at t=2.40s

1. What are the kinematic equations used to calculate position and velocity at t=2.40s?

The kinematic equations used to calculate position and velocity at t=2.40s are the following:

Position: x = x0 + v0t + 1/2at2

Velocity: v = v0 + at

Where x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

2. What is the meaning of each variable in the kinematic equations?

The variables in the kinematic equations have the following meaning:

x: final position

x0: initial position

v: final velocity

v0: initial velocity

a: acceleration

t: time

3. How do I know which kinematic equation to use?

The kinematic equation to use depends on what information you have and what information you are trying to find. If you know the initial and final positions, initial velocity, and acceleration, you can use the position equation. If you know the initial velocity, acceleration, and time, you can use the velocity equation.

4. Can the kinematic equations be used for objects with constant acceleration?

Yes, the kinematic equations can be used for objects with constant acceleration. However, they may not accurately represent the motion of objects with non-constant acceleration.

5. How do I solve for a missing variable in the kinematic equations?

If you have three known variables, you can rearrange the equation to solve for the missing variable. For example, if you have the initial and final positions and the acceleration, you can use the position equation to solve for the initial velocity. If you have the initial velocity, acceleration, and time, you can use the velocity equation to solve for the final velocity.

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