Fundamental kinematics question

In summary, the homework statement states that a constant force of 8.0N is exerted for 4.0s on a 16-kg object initially at rest. The change in speed of the object will be 2 m/s at the end of the 4th second.
  • #1
6
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Homework Statement


I seem to have a fundamental misunderstanding of the kinematic principles in this question.

A constant force of 8.0N is exerted for 4.0s on a 16-kg object initially at rest. What will the change in speed of this object be?

Homework Equations


F = ma
Δx = v0t + 1/2 at2
Δv = v0 +at
Δv = Δx/Δt
Δv = v0 + aΔt

The Attempt at a Solution


F = ma

thus

a = (8.0N) / (16.0 kg) = .5 m/s^2

then

Δx = (0 m/s)(4 s) + 1/2(.5 m/s^2)(4 s)^2 = 4 m

if

Δv = Δx/Δt

then

Δv = (4 m) / (4 s) = 1 m/s

but using Δv = aΔt

Δv = (.5 m/s^2)(4 s) = 2 m/s

I don't understand why I would have two conflicting answers there. Just curious if anyone might have some insight on why that would be. Thanks!
 
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  • #2
[itex] \frac{ \triangle x}{\triangle t} [/itex] is an average velocity over the 4 second interval, not the instanteous velocity of object at the end of the 4 seconds.
 
  • #3
abram said:
Δv = Δx/Δt
What's this equation mean to you?
 
  • #4
Bystander said:
What's this equation mean to you?

There are various interpretations. One is that if velocity is constant with respect to time then it is equal to the change of distance over a time interval divided by the length of that time interval. Thinking of the equation as a way to remember some calculus, you could think of the meaning: "Instanteous velocity at a given time is equal to instantaneous rate of change of distance with respect to time at that instant". The way you used the equation was to compute an average velocity.
 
  • #5
I was asking the OP, not you, as a hint, while we posted simultaneously.
 
  • #6
Bystander said:
I was asking the OP, not you, as a hint, while we posted simultaneously.

I should have noticed that you weren't the OP.
 
  • #7
abram said:

Homework Statement


I seem to have a fundamental misunderstanding of the kinematic principles in this question.

A constant force of 8.0N is exerted for 4.0s on a 16-kg object initially at rest. What will the change in speed of this object be?

Homework Equations


F = ma
Δx = v0t + 1/2 at2
Δv = v0 +at
Δv = Δx/Δt

Δv = v0 + aΔt


Your relevant equations are not correct. It is uniformly accelerating motion, that means v = v0 +at, where v0 is the velocity at t=0, or change of velocity in the interval Δt is Δv=aΔt.
The displacement Δx= v0t + 1/2 at2 is correct.
 
  • #8
Thanks for the quick reply. I suspected that at first as well. I then thought that if 2 m/s was the final velocity, then

Δv = (2 m/s) - (0 m/s) = 2 m/s

I just couldn't see how 1 m/s could be the average, but I'll play around with it some more and see how that happens. Thanks again.
 
  • #9
For v = v0 + at, that would play out as:

v = (0 m/s) + (.5 m/s2)(4 s) = 2 m/s

I'm confusing average velocity with change in velocity. I new there was something. Thanks again folks!
 
  • #10
For the average of two velocities, it would be something like:

vavg = (2 m/s - 0 m/s)/2 = 1 m/s
 
  • #11
abram said:
For v = v0 + at, that would play out as:

v = (0 m/s) + (.5 m/s2)(4 s) = 2 m/s
The change of speed was asked.
Yes, the speed will be 2 m/s at the end of the 4th second. As it was zero initially, so the change of speed is 2 m/s.

The average velocity is defined for a time interval Δt as displacement Δx over time Δt: vaverage=Δx/ Δt.
You have the formula for Δx=voΔt+a/2 Δt2. As vo = 0, Δx=a/2 Δt2, and vaverage=Δx/ Δt=a/2 Δt.
There is a other formula for the displacement in case the velocity changes from v1 to v2 in time Δt: ##\Delta x = \frac{v_1+v_2}{2}\Delta t##.
In case of uniformly accelerating motion, the average of the velocity in a time interval is also the average (mean) of the initial and final velocities.
 
  • #12
ehild said:
The change of speed was asked.
Yes, the speed will be 2 m/s at the end of the 4th second. As it was zero initially, so the change of speed is 2 m/s.

The average velocity is defined for a time interval Δt as displacement Δx over time Δt: vaverage=Δx/ Δt.
You have the formula for Δx=voΔt+a/2 Δt2. As vo = 0, Δx=a/2 Δt2, and vaverage=Δx/ Δt=a/2 Δt.
There is a other formula for the displacement in case the velocity changes from v1 to v2 in time Δt: ##\Delta x = \frac{v_1+v_2}{2}\Delta t##.
In case of uniformly accelerating motion, the average of the velocity in a time interval is also the average (mean) of the initial and final velocities.

Roger that. Good to know!
 

What is fundamental kinematics?

Fundamental kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It is concerned with concepts such as displacement, velocity, and acceleration.

What are the basic equations of kinematics?

The basic equations of kinematics are the equations of motion, which are:
- Position (x) = Initial position (x0) + Velocity (v) x Time (t)
- Velocity (v) = Initial velocity (v0) + Acceleration (a) x Time (t)
- Displacement (Δx) = Initial velocity (v0) x Time (t) + 1/2 x Acceleration (a) x Time (t)^2
- Velocity (v)^2 = Initial velocity (v0)^2 + 2 x Acceleration (a) x Displacement (Δx)

What is the difference between distance and displacement?

Distance is the total length traveled by an object, while displacement is the shortest distance between the initial and final positions of the object. Distance is a scalar quantity, while displacement is a vector quantity with both magnitude and direction.

What is the difference between speed and velocity?

Speed is the rate at which an object covers distance, while velocity is the rate at which an object changes its position. Speed is a scalar quantity, while velocity is a vector quantity. Speed does not consider direction, while velocity does.

How is acceleration related to velocity?

Acceleration is the rate of change of velocity. This means that acceleration describes how an object's velocity changes over time. An object can have a constant velocity but still have an acceleration if its direction of travel changes.

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