# How Do Velocity and Acceleration Change Under Unbalanced Forces?

• sparklegemini
In summary, for the conversation about the forces acting on a mass of 1.0 kg, F1 is constant at 1.0 N to the left in the negative x-direction and F2(t) varies with a magnitude of 2.0 N in the positive x-direction for the time interval 0.0 s ≤ t ≤ 5.0 s, dropping to zero after 5.0 s. The mass is initially at rest at t = 0.0 s. The acceleration versus time graph for the mass during the time interval 0.0 s ≤ t ≤ 15.0 s is constant at a = +1 m/s/s for the first 5 seconds and then drops to 0 for the
sparklegemini
HW Template missing as it was moved from another forum
1. problem
Two forces, F1 and F2(t), act on a mass m = 1.0 kg. Force 1 F1 is to the left in the negative x-direction and force F2(t) is to the right in the positive x-direction. F1 is constant with a magnitude of 1.0 N whereas F2(t) has a magnitude of 2.0 N for the time interval 0.0 s ≤ t ≤ 5.0 s and drops to zero after 5.0 s. The mass is at rest at time t = 0.0 s.

a) Sketch the acceleration versus time graph for the time interval 0.0 s ≤ t ≤ 15.0 s for the mass m.
b) sketch the v(t) for the same time as question a.

2. attemps:
I am confusing about the time t=0s, when the object is at rest, that means a=0 and v=0? no unbalanced forces?
Since f2=2N and f1 =1N during the interval time [0,5]. netforce would be 1N. Because F2(t)=2N during [0,5] and it comes to 0 after 5s. for example t=6s.7.,,,. I believe that i shoud calculate the friction force cos It will be 1N in magnitude and the same negative direction as F1.
Now when t>5s, t=6s, Fnet is now missing F2. Will the mass start acceleration? although I have considered the FFric.? If it does, a=-1m/s/s. so velocity will be -1 at t=6, hence -2,-3... at t=7,8... while a is now constant and has the value of -1?
and if there is friction consideration, fnet =0 cos friction force will resist the tend to accelerate by F1

sparklegemini said:
I am confusing about the time t=0s, when the object is at rest, that means a=0 and v=0? no unbalanced forces?
no. "At rest" means the speed is instantaneously zero.

Since f2=2N and f1 =1N during the interval time [0,5]. netforce would be 1N. Because F2(t)=2N during [0,5] and it comes to 0 after 5s.
the net force is different for different times. Force is a vector, don't forget to say which direction the net force points in.

...for example t=6s.7.,,,. I believe that i shoud calculate the friction force cos It will be 1N in magnitude and the same negative direction as F1.
... do not assume what is not in the problem statement. Friction is not mentioned anywhere so what makes you think it is 1N?

Now when t>5s, t=6s, Fnet is now missing F2. Will the mass start acceleration? although I have considered the FFric.?
... does it have an unbalanced force acting on it?

If it does, a=-1m/s/s. so velocity will be -1 at t=6, hence -2,-3... at t=7,8... while a is now constant and has the value of -1?
and if there is friction consideration, fnet =0 cos friction force will resist the tend to accelerate by F1
... no friction is mentioned, acceleration is constant inside both time intervals of interest.

Thank for your guiding Simon, So I am still in trouble with the time t=0s, does F2 has magnitude of 2N and right ward direction?
if mass is at rest means it has V instantaneously =0, that means since F2 applied forced, There is unbalanced force act on mass, Fnet=F2-F1, a=1 and in positive direction. Velocity has changed to 0 m/s when F2 started?

before we consider the F2 started. time before we call it 0s, Object is moving with only Fnet=F1=ma1, a1=-1, negative positions ? again, velocity changed to 0 and changed also direction to positive direction when F2 applied?

Thank for your guiding Simon, So I am still in trouble with the time t=0s, does F2 has magnitude of 2N and right ward direction?
Consider: what dfference does it make to your graph?

if mass is at rest means it has V instantaneously =0, that means since F2 applied forced, There is unbalanced force act on mass, Fnet=F2-F1, a=1 and in positive direction. Velocity has changed to 0 m/s when F2 started?
velocity is equal to 0 at t=0s ... it is not changed to that.
We are not told what happened at t<0s so don't assume.

before we consider the F2 started. time before we call it 0s,
When does the problem say that F2 started?

Object is moving with only Fnet=F1=ma1, a1=-1, negative positions ? again, velocity changed to 0 and changed also direction to positive direction when F2 applied?
At t=0, what is the position of the object? What is the net force? If you don't answer questions I cannot help you.
This problem is very straight forward and you are over-thinking it.

Last edited:
Simon Bridge said:
Consider: what dfference does it make to your graph?

velocity is equal to 0 at t=0s ... it is not changed to that.
We are not told what happened at t<0s so don't assume.When does the problem say that F2 started?

At t=0, what is the position of the object? What is the net force? If you don't answer questions I cannot help you.
This problem is very straight forward and you are over-thinking it.
Hum, I think that at t=0, the object is at rest, Velocity equal 0 and Netforce is 1N following F2 direction, with the acceleration of 1m/s/s ? right

Directions are "left" and "right" or "x direction".
So at t=0, v(0)=0 a(0)=F(0)/m and F(0)=F2-F1 in the +x direction. ... well done.

I used brackets since v a and F are functions of time... so is x. v(0)=v(t=0) see?

In fact. for ##0\leq t\leq5## (seconds) ##F=+1##N ... the ##+## indicates the ##+x## direction.

Take this one step at a time:
Using the same notation, what is the acceleration for ##0\leq t \leq 5##s?

Simon Bridge said:
Directions are "left" and "right" or "x direction".
So at t=0, v(0)=0 a(0)=F(0)/m and F(0)=F2-F1 in the +x direction. ... well done.

I used brackets since v a and F are functions of time... so is x. v(0)=v(t=0) see?

In fact. for ##0\leq t\leq5## (seconds) ##F=+1##N ... the ##+## indicates the ##+x## direction.

Take this one step at a time:
Using the same notation, what is the acceleration for ##0\leq t \leq 5##s?
Hi, since Fnet unchanges during interval time above, a=Fnet/m=+1, so what about velocity ? does it follow the equation v=vo+ a.t?
v(0)=0, v(5)= 0+1x5=5? I wonder why the answersheet covers only maximum of +2 of velocity ( in vertical line) for the graph to be drawed

After 5s, a(t>5)=Fnet(t>5)/m Fnet now is -1 since Only F1 act on the mass. negative direction. so a=-1 negative direction.?
again, how about velocity, if it was previously 5 m/s, now with the acceleration -1, velocity changes direction and its value decreases by 1 m/s per second?

sparklegemini said:
Hi, since Fnet unchanges during interval time above, a=Fnet/m=+1, so what about velocity ? does it follow the equation v=vo+ a.t?
v(0)=0, v(5)= 0+1x5=5? I wonder why the answersheet covers only maximum of +2 of velocity ( in vertical line) for the graph to be drawed

After 5s, a(t>5)=Fnet(t>5)/m Fnet now is -1 since Only F1 act on the mass. negative direction. so a=-1 negative direction.?
again, how about velocity, if it was previously 5 m/s, now with the acceleration -1, velocity changes direction and its value decreases by 1 m/s per second?
I have got some ideas, can it is possible to assume that only during 5 seconds, a =+1, so velocity changes to +1. and from 5-10, it drops to 0 since a=-1 and from 10-15s, a remains -1 and velocity is now -1., if 15-20s, velocity -2...

Hi, since Fnet unchanges during interval time above, a=Fnet/m=+1, so what about velocity ? does it follow the equation v=vo+ a.t?
v(0)=0, v(5)= 0+1x5=5?
... well done.
For the interval in question, v(t)=t so v(5)=5m/s

I wonder why the answersheet covers only maximum of +2 of velocity ( in vertical line) for the graph to be draw[n]
No idea - I didn't set the question.
Velocity should be on the vertical axis and the line of constant velocity should be horizontal. You will need to extend the graph or relabel the axes (I'd suggest the latter), but check you have read the question properly ... maybe you missed out something. It could also be an edit of an earlier question and the editor forgot to change the graph axes values.

After 5s, a(t>5)=Fnet(t>5)/m Fnet now is -1 since Only F1 act on the mass. negative direction. so a=-1 negative direction.?
Well done ... so:
if it was previously 5 m/s, now with the acceleration -1, velocity changes direction and its value decreases by 1 m/s per second?
... correct: for ##t>5## the velocity is given by ##v(t>5)=v(5)-t##. The graph just decreases with the same slope that it increases before.

Taking things one step at a time stops different stages of the problem from confusing you.
You can test your understanding by working out how long before the object returns to x=0 if you like.

While I was typing you wrote:
I have got some ideas, can it is possible to assume that only during 5 seconds, a =+1, so velocity changes to +1. and from 5-10, it drops to 0 since a=-1 and from 10-15s, a remains -1 and velocity is now -1., if 15-20s, velocity -2...
... and you were doing so well too :(
You cannot just pluck numbers out of the air - you have to use physics.
If the acceleration is 1m/s2 then v=1m/s after 1s has passed.

Simon Bridge said:
... well done.
For the interval in question, v(t)=t so v(5)=5m/s

No idea - I didn't set the question.
Velocity should be on the vertical axis and the line of constant velocity should be horizontal. You will need to extend the graph or relabel the axes (I'd suggest the latter), but check you have read the question properly ... maybe you missed out something. It could also be an edit of an earlier question and the editor forgot to change the graph axes values.

Well done ... so: ... correct: for ##t>5## the velocity is given by ##v(t>5)=v(5)-t##. The graph just decreases with the same slope that it increases before.

Taking things one step at a time stops different stages of the problem from confusing you.
You can test your understanding by working out how long before the object returns to x=0 if you like.

While I was typing you wrote:... and you were doing so well too :(
You cannot just pluck numbers out of the air - you have to use physics.
If the acceleration is 1m/s2 then v=1m/s after 1s has passed.
oh yes, Thank a lot. I just think crazy out of a box when velocity was too high than the graph expected to be drawn.
I attached the area for the graphs.
they would draw the wrong coordinate for velocity

#### Attachments

• image.jpg
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Like I said - just relabel the axes after checking that you have read the question correctly.
The top one (that goes to 6) looks like it has the correct range for the v(t) graph and the bottom one looks correct for a(t)... so maybe they just got them switched.

Simon Bridge said:
Like I said - just relabel the axes after checking that you have read the question correctly.
The top one (that goes to 6) looks like it has the correct range for the v(t) graph and the bottom one looks correct for a(t)... so maybe they just got them switched.
yes, I got it. thank so much for your help Simo

## What is a v(t) and a(t) graph?

A v(t) and a(t) graph is a graphical representation of an object's velocity (v) and acceleration (a) over time (t). It shows how an object's velocity and acceleration change over time, providing valuable information about its motion.

## How do you draw a v(t) and a(t) graph?

To draw a v(t) and a(t) graph, you first need to collect data on an object's position at different points in time. Then, plot the position data on the y-axis and time data on the x-axis. The slope of the resulting curve will give you the object's velocity, while the slope of the velocity curve will give you the object's acceleration.

## What does a flat line on a v(t) and a(t) graph represent?

A flat line on a v(t) and a(t) graph represents a constant velocity or acceleration. This means that the object is not speeding up or slowing down, but maintaining a steady speed or rate of change.

## How can you tell if an object is accelerating on a v(t) and a(t) graph?

If the slope of the velocity curve is increasing or decreasing, it means that the object is accelerating. This is because acceleration is defined as the rate of change of velocity, so a changing slope indicates a changing velocity.

## Can you determine the displacement of an object from a v(t) and a(t) graph?

Yes, you can determine the displacement of an object by finding the area under the velocity curve on a v(t) and a(t) graph. This is because displacement is equal to the integral of velocity with respect to time. So, the area under the curve represents the change in position over time.

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