Velocity due to known acceleration

In summary, the conversation involved a person trying to solve a problem by finding the acceleration using F = ma, then integrating the graph to find the velocity at t = 10 s. However, there were some mistakes in the person's approach and it was suggested to use impulse instead. The conversation concluded with the person successfully solving the problem using both methods.
  • #1
Anders Hijul
6
2
Homework Statement
A particle with a mass 𝑚 = 10.0kg has the time 𝑡 = 0 s velocity 𝑣0 = 10.0m / s. The particle is affected by a force 𝐹 that depends on the time as shown in the figure below.

What is the velocity of the particle at time 𝑡 = 10 s?
Relevant Equations
F = ma
I have some troubles with this relatively simple problem My idea was to find the acceleration by F = ma and then integrate the graph and then find the velocity to t = 10 s + start velocity

The graph will be
- 2x

And integrated
-x¨^2

But this seems wrong

Thanks in regards
 

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  • #2
What is ##x## here? You are given ##F## as a function of time.
 
  • #3
Sorry my mistake I meant t instead of x
 
  • #4
Anders Hijul said:
Sorry my mistake I meant t instead of x
Can you write down what you are actually doing?
 
  • #5
I meant that the graph would be
20-2t for t = 5 to 5 = 10 s

But integrated the graph is
20t - t^2

But I think I am doing something wrong before this
 
  • #6
Anders Hijul said:
I meant that the graph would be
20-2t for t = 5 to 5 = 10 s

But integrated the graph is
20t - t^2

But I think I am doing something wrong before this
First, you need some units there. ##20t - 2t^2## is not a valid physical quantity.

Second, I thought you were trying to calculate the acceleration as a function of time?

By the way, have you ever heard the term impulse?
 
  • #7
I think I maybe haven't explained myself correctly

I want to find the velocity of the object to t = 10 s

When I have the force applied to the object over time, I can calculate the acceleration from F = ma, so I get
a = 1 m/s^2 for t = 0 to t = 5
a = 0,8 m/s^2 for t = 6
a = 0,6 m/s^2 for t = 4

And so on My idea was to simply integrate the graph of the acceleration to get the graph of the velocity and then find the velocity to t = 10 s plus the start velocity (for t = 0 to t = 5 s)

Hope that clears it up

Yes I have heard of impulse, but I cannot see how I can apply that here

Thanks
 
  • #8
Anders Hijul said:
I think I maybe haven't explained myself correctly

I want to find the velocity of the object to t = 10 s

When I have the force applied to the object over time, I can calculate the acceleration from F = ma, so I get
a = 1 m/s^2 for t = 0 to t = 5
a = 0,8 m/s^2 for t = 6
a = 0,6 m/s^2 for t = 4

And so on My idea was to simply integrate the graph of the acceleration to get the graph of the velocity and then find the velocity to t = 10 s plus the start velocity (for t = 0 to t = 5 s)

What's the velocity after ##5## seconds?

What's your plan for integrating acceleration over time from ##5## to ##10## seconds? Can you express the acceleration as a function of time?
 
  • #9
PS it's easier to calculate the total impulse!
 
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Likes PSN03
  • #10
Anders Hijul said:
I meant that the graph would be
20-2t for t = 5 to 5 = 10 s

But integrated the graph is
20t - t^2

But I think I am doing something wrong before this
I think you have some mistakes in your integration.
You should try applying impulse here as @PeroK said. It will be much more easier
For your information F*t=impulse=change in momentum
 
  • #11
Thanks PeroK and PSN03

I think I got it!

So the impulse is calculated from force applied, which is 75 N (area under the graph). So for 10 s we have
J = F*t = 75 N*10 s = 75 N s

So we have
J = pf-pi
75 N = 10 kg*v -> v = 7,5 m/s

So the final velocity after 10 s is 10 + 7,5 = 17,5 m/s

It funny with Physics. When you figure something out, it seems so simple :wink:
 
  • #12
Anders Hijul said:
Thanks PeroK and PSN03

I think I got it!

So the impulse is calculated from force applied, which is 75 N (area under the graph). So for 10 s we have
J = F*t = 75 N*10 s = 75 N s

So we have
J = pf-pi
75 N = 10 kg*v -> v = 7,5 m/s

So the final velocity after 10 s is 10 + 7,5 = 17,5 m/s

It funny with Physics. When you figure something out, it seems so simple :wink:
The area under a Force/time graph is impulse:
$$J = (10N \times 5s) + \frac 1 2 (10N \times 5s) = 75 Ns$$
I would also have written:
$$\Delta v = \frac J m = 7.5 m/s$$
 
  • #13
That make sense

Thank you!
 
  • #14
Anders Hijul said:
Thanks PeroK and PSN03

I think I got it!

So the impulse is calculated from force applied, which is 75 N (area under the graph). So for 10 s we have
J = F*t = 75 N*10 s = 75 N s

So we have
J = pf-pi
75 N = 10 kg*v -> v = 7,5 m/s

So the final velocity after 10 s is 10 + 7,5 = 17,5 m/s

It funny with Physics. When you figure something out, it seems so simple :wink:
You can also apply your integration method as the slope of the curve will be
F/t=m
Ma/t=m
(M*dv/dt)/t=m
Mdv=m*t dt
Now just integrate this thing
PS: here M is mass of body and m is slope of the line
 

Related to Velocity due to known acceleration

1. What is the formula for calculating velocity due to known acceleration?

The formula for calculating velocity due to known acceleration is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. How is acceleration related to velocity?

Acceleration is the rate of change of velocity over time. This means that when an object experiences acceleration, its velocity changes by a certain amount in a certain direction over a certain period of time.

3. What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time taken, while instantaneous velocity is the velocity of an object at a specific moment in time. Average velocity gives an overall picture of an object's motion, while instantaneous velocity shows the object's velocity at a specific point in time.

4. Can an object have a constant acceleration and varying velocity?

Yes, an object can have a constant acceleration and varying velocity. This is because acceleration is the rate of change of velocity, so as long as the velocity is changing at a constant rate, the acceleration remains constant.

5. How does the direction of acceleration affect an object's velocity?

The direction of acceleration affects an object's velocity by changing its direction. For example, if an object is accelerating in the opposite direction of its velocity, its velocity will decrease. On the other hand, if the acceleration is in the same direction as the velocity, the velocity will increase.

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