What is the acceleration of an object moving with a constant acceleration?

In summary, the equation used to calculate the magnitude of acceleration for an object moving with constant acceleration in a straight line is d=v_0t+\frac{at^2}{2}, where d is the distance, v_0 is the initial velocity, t is the time interval, and a is the acceleration. By substituting the given data for position and time, the magnitude of acceleration for the object at t=38.60 s is calculated to be 2.60 m/s^2.
  • #1
CoverlessTech
4
0

Homework Statement



An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.

Time (s) Position, (m)
37.10 7.600
38.60 14.425
40.10 27.100

Calculate the magnitude of the acceleration at t=38.60 s.


Homework Equations



a = (vf - vo) / t
d = vot + 1/2 at2
vf2 = vo2 + 2ad
(ether or really)

The Attempt at a Solution



I've tried everything, I've tried getting the distances between the 2 each time and using 1.5 seconds and getting velocity, I've tried using those to get the velocity, nothing is working. and this is something I should be able to do. Any help would be good, even just a point in the right direction.
 
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  • #2
Welcome to the Physics Forums!

In the equation [tex]d=v_0t+\frac{at^2}{2}[/tex]
[tex]d[/tex] is a distance. But in the statement of the problem it says that the position of the object is measured...so you can't replace d with the data for the position.
But if you write the distance d as the difference between the position (say [tex]x[/tex]) measured at a time and the initial position (say [tex]x_0[/tex]) and replace in the equation, then you have:

[tex]x-x_0=v_0t+\frac{at^2}{2}[/tex]

Now all you have to do is replace the 3 sets of data in the equation above and you have a system of 3 equations with 3 unknowns (one of them being the acceleration).
 
Last edited:
  • #3
Thanks but I just got it. Before I was using 3 seconds when I was suppose to be using 1.5s, what a silly mistake.

<I>d1</I> = 14.425-7.600 = 6.825 m / 1.5s(38.60 - 37.10) = <i>v1</I> = 4.55 m/s
<I>d2</I> = 27.100-14.425 = 12.675 m / 1.5s = <i>v2</I> = 8.45 m/s
<T> = 1.5s

V1-V2/T = 2.60 m/s^2
 
  • #4
CoverlessTech said:
Thanks but I just got it. Before I was using 3 seconds when I was suppose to be using 1.5s, what a silly mistake.

<I>d1</I> = 14.425-7.600 = 6.825 m / 1.5s(38.60 - 37.10) = <i>v1</I> = 4.55 m/s
<I>d2</I> = 27.100-14.425 = 12.675 m / 1.5s = <i>v2</I> = 8.45 m/s
<T> = 1.5s

V1-V2/T = 2.60 m/s^2

That is also correct.
 

1. What is Kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the cause of the motion.

2. What are the three basic quantities used in kinematics?

The three basic quantities used in kinematics are displacement, velocity, and acceleration. Displacement is the change in position of an object, velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

3. How is kinematics different from dynamics?

Kinematics deals with the description of motion, while dynamics deals with the causes of motion, such as forces and energy.

4. How is kinematics used in real life?

Kinematics is used in various fields such as engineering, robotics, and sports. It helps in designing and analyzing the motion of objects and predicting their behavior.

5. What are the most common equations used in kinematics?

The most common equations used in kinematics are the equations of motion, which include the equations for displacement, velocity, and acceleration. They are often referred to as the Big Three equations: s = ut + 0.5at², v = u + at, and v² = u² + 2as.

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