Velocity due to known acceleration

[Anders Hijul]

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

 

[PeroK]

What is $x$ here? You are given $F$ as a function of time.

 

[Anders Hijul]

Sorry my mistake I meant t instead of x

 

[PeroK]

Sorry my mistake I meant t instead of x

Can you write down what you are actually doing?

 

[Anders Hijul]

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

 

[PeroK]

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?

 

[Anders Hijul]

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

 

[PeroK]

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?

 

[PeroK]

PS it's easier to calculate the total impulse!

 

[PSN03]

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

 

[Anders Hijul]

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

 

[PeroK]

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

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:
$$\Delta v = \frac J m = 7.5 m/s$$

 

[Anders Hijul]

That make sense

Thank you!

 

[PSN03]

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

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