Fundamental kinematics question

[abram]

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!

 

[Stephen Tashi]

$\frac{ \triangle x}{\triangle t}$ is an average velocity over the 4 second interval, not the instanteous velocity of object at the end of the 4 seconds.

 

[Bystander]

Δv = Δx/Δt

What's this equation mean to you?

 

[Stephen Tashi]

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.

 

[Bystander]

I was asking the OP, not you, as a hint, while we posted simultaneously.

 

[Stephen Tashi]

I was asking the OP, not you, as a hint, while we posted simultaneously.

I should have noticed that you weren't the OP.

 

[ehild]

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.

 

[abram]

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.

 

[abram]

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!

 

[abram]

For the average of two velocities, it would be something like:

v_avg = (2 m/s - 0 m/s)/2 = 1 m/s

 

[ehild]

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: v_average=Δx/ Δt.
You have the formula for Δx=voΔt+a/2 Δt². As vo = 0, Δx=a/2 Δt², and v_average=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.

 

[abram]

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: v_average=Δx/ Δt.
You have the formula for Δx=voΔt+a/2 Δt². As vo = 0, Δx=a/2 Δt², and v_average=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!