What is the speed of the bicycle?

[brochesspro]

How about https://drive.google.com/file/d/19XcGYjYZOe_M2mPN9nujJ7JuldeHovlk/view?usp=sharing?

Can you open this link?

 

[PeroK]

Can you open this link?

You need to post some Latex or, at least, an image of your work.

 

[brochesspro]

It has an image, if you can open it.

 

[PeroK]

It has an image, if you can open it.

You need to post your solution in this thread; not a link to an external site.

 

[brochesspro]

You need to post your solution in this thread; not a link to an external site.

Ok, but is that a rule?

 

[PeroK]

Ok, but is that a rule?

Yes.

 

[brochesspro]

 

[brochesspro]

Then what is the "Insert link" button used for?

 

[PeroK]

View attachment 289367View attachment 289368View attachment 289369

There are a few mistakes there. $t_2 = 3s$ is correct. The calculation of $x_{t_2}$ has gone wrong somewhere. You added an extra $12s$ for some reason(?).

 

[brochesspro]

I added the 12 seconds by mistake. I should have added 1 second for the phase where the speed is constant.

 

[brochesspro]

There are a few mistakes there. $t_2 = 3s$ is correct. The calculation of $x_{t_2}$ has gone wrong somewhere. You added an extra $12s$ for some reason(?).

I too feel I have gone wrong in that part.

 

[PeroK]

I added the 12 seconds by mistake. I should have added 1 second for the phase where the speed is constant.

The way you did it, $t_2$ is the final time. You don't need to add anything.

I too feel I have gone wrong in that part.

I don't see the error immediately - but I used $T$ to keep things simple.

 

[brochesspro]

Oh, I see, so I did a mistake in something basic.

 

[brochesspro]

I think I neglected the constant $C$ while integrating for the first time for finding $x_{t_2}$.

 

[PeroK]

Oh, I see, so I did a mistake in something basic.

You have $a(t) = -6(t - 1)$ (let's do the maths without units). So: $$v(t) = -3t^2 + 6t + k$$ where $k$ is the constant of integration. We know that $v(1) = 12$ (note that we do not have $v(0) = 12$). So:
$$k = 12 + 3 - 6 = 9$$ and $$v(t) = -3t^2 + 6t + 9$$ That gives $v(3) = 0$. Integrating again:
$$x(t) = -t^3 +3t^2 +9t + l$$
I think that's where you went wrong, by losing the $9t$ term.

 

[brochesspro]

Never mind, I saw your post.

 

[PeroK]

Should I take $C = 12 m$ as it is the initial velocity of the body?

The constant of integration in the equation for $x(t)$ doesn't matter, as you are calculating the distance traveled between $t = 1$ and $t =3$.

 

[brochesspro]

After integration, I got $x_{t_2}-x_{t_1}=16$ after solving and after substituting I got $x_{t_2}=28m$ and thus the speed of bicycle is $9.3 m/s$ as per the question.

 

[PeroK]

After integration, I got $x_{t_2}-x_{t_1}=16$ after solving and after substituting I got $x_{t_2}=28m$ and thus the speed of bicycle is $9.3 m/s$ as per the question.

The bicycle is another $17m$ ahead to begin with. It must be Mark Cavendish after all!

 

[brochesspro]

Oh yeah, so I get the average velocity as $15 m/s$.

 

[brochesspro]

Thanks a lot.
So, how do I close this thread?

 

[PeroK]

Thanks a lot.
So, how do I close this thread?

It doesn't need to get closed. It can stay as it is.

 

[brochesspro]

Oh, I see. Thanks again.