Meeting of two bodies (cinematic)

[A13235378]

Homework Statement

Two pieces of furniture, A and B, run along the same line, in the same direction, in such a way that,
instant 𝑡 = 0.00 𝑠 the distance between them is 10.0 𝑚. The graphs of your speeds are
shown in the figure. It is known that the furniture passes by each other in a certain moment 𝑡𝐸> 0, where the speed of B in relation to that of A has a certain value 𝑣𝐵𝐴. We can conclude that:

Relevant Equations

V=Vo+at
S=So+vt+at^2/2

First, I calculated the acceleration of A by the slope:

-0.5m / s ^ 2

in t=4/s

V=Vo+at (To bodie A)
V=4-4.0,5
V=2m/s

With that, I was able to find the acceleration of B:

0,5m/s^2SA= 10 +4t - t^2/4
SB= t^2/4

I equaled the two, finding t = 10s

so, Va=-1m/s and Vb = 5m/s --> Vb,a = 6 m/s

BUT...

it's a matter of multiple choice and the answer is:

The problem as proposed has no solution.Why?

I know that I assumed that body A would be 10 m ahead, because only then would there be a solution to the second degree equation.

 

[BvU]

HI,

Could you read the problem statement as posted

Homework Statement:: Two pieces of furniture, A and B, run along the same line, in the same direction, in such a way that,
instant 𝑡 = 0.00 𝑠 the distance between them is 10.0 𝑚. The graphs of your speeds are
shown in the figure. It is known that the furniture passes by each other in a certain moment 𝑡𝐸> 0, in the what the speed of B in relation to that of A has a certain value 𝑣𝐵𝐴. We can conclude that:
Relevant Equations:: V=Vo+at
S=So+vt+at^2/2

and fix the mistakes ?

 

[Steve4Physics]

Homework Statement:: Two pieces of furniture, A and B, run along the same line, in the same direction, in such a way that,
instant 𝑡 = 0.00 𝑠 the distance between them is 10.0 𝑚. The graphs of your speeds are
shown in the figure. It is known that the furniture passes by each other in a certain moment 𝑡𝐸> 0, where the speed of B in relation to that of A has a certain value 𝑣𝐵𝐴. We can conclude that:
.

What you have written is hard to follow. I’m guessing that English is not your first language. But the problems are not entirely language related. It will help if you check your post carefully. For example:

1. If the question uses a certain symbol, you should use the same symbol in your working (e.g. $t_E$).

2. Using a comma as a ‘decimal separator’ can be confusing. The convention in most English-speaking countries is to use a full-stop (‘point’). So ½ is written as 0.5 not 0,5 for example.
Take a look at this:https://brilliantmaps.com/decimals/
Since this is an English-speaking forum, you might prefer to write (for example):
V = 4 – 4*0.5
rather than
V=4-4.0,5

3. Spelling mistakes indicate that you haven’t checked what you are posting. E.g. ‘cinematic’ should be ‘kinematics’ and ‘bodie’ should be ‘body’.

4. Check what you have written makes sense. E.g. ‘in t=4/s’ means nothing. Maybe you mean ‘When t=4s’.

5. You haven’t actually stated the question! But, from your working, I’d guess the question is to find $V_{BA}$ (velocity of B relative to A) at time $t=t_E$.

6. You haven’t given the list of the alternative answers. Mistakes can happen - an answer indicated as ‘correct’ might be wrong. So it helps to see the full list.

7. If you can learn how to use simple Latex to format equations, your post would be more readable.

There are probably other improvements but the above are the ones I easily spotted.

If you can improve your post so it’s easy to read/understand, I’m sure you will get some help.

 

[Steve4Physics]

SA= 10 +4t - t^2/4
SB= t^2/4

I equaled the two, finding t = 10s

The time can’t be 10s because the graph finishes at 8s. You don't know what happens after 8s.

You have incorrectly used signs/sign-conventions. You have to get the signs of velocity and acceleration correct according to whether an object is moving right (+x) or left (-x).

Remember a negative acceleration means the *direction* of acceleration is negative. This can result in speeding-up or slowing-down.

Let's say:

A starts at $x_0$ = 0m with u = 4m/s and a = -0.5$m/s^2$
That means A is initially moving right. Because acceleration is negative (to the left), the speed reduces while A moves to the right.
At time t: $x_A = x_0 + ut + ½at^2 = 0 + 4t + ½(-0.5)t^2$
$x_A = 4t - 0.25t^2$

B starts at $x_0$ = 10m with u = 0m/s and a = -0.5$m/s²$.
That means B is initially stationary but because acceleration is negative, B moves left with increasing speed.
At time t: $x_B = x_0 + ut + ½at^2 = 10 + 0t + ½(-0.25)t^2$
$x_B = 10 - 0.25t^2$

When A and B meet at time $t_E$ then $x_A=x_B$
$4t_E - 0.25t_E^2 = 10 – 0.25t_E^2$
$t_E = 10/4 = 2.5m/s$

Note the '$½at^2$’ terms have cancelled. That's because both objects have the same (negative) acceleration: the speed of approach between A and B is always 4m/s. It's an easy relative velocity problem, the quick way to find $t_E$ is to use:
$t_E = \frac {initial-separation}{speed-of-approach} = \frac{10m}{4m/s}$

$V_{BA}$ can then be found (assuming that is the question!).

[Edited several times to correct typo's]

 

[Steve4Physics]

My answer in post #4 is wrong. I thought A and B were moving in opposite directions but the question says they are moving in the *same* direction.

So the correct working is:

A starts at $x_0$ = 0m with u = 4m/s and a = -0.5$m/s^2$
$x_A = x_0 + ut + ½at^2 = 0 + 4t + ½(-0.5)t^2$
$x_A = 4t - 0.25t^2$

B starts at $x_0$ = 10m with u = 0m/s and a = 0.5$m/s²$.
$x_B = x_0 + ut + ½at^2 = 10 + 0t + ½(0.5)t^2$
$x_B = 10 + 0.25t^2$

When A and B meet at time $t_E$ then $x_A=x_B$
$4t_E - 0.25t_E^2 = 10 + 0.25t_E^2$
$0.5t_E^2 – 4t_E + 10 = 0$
$t_E^2 – 8t_E + 20 = 0$
This has no (real) solution so A never catches up to B.

It goes to show how important it is to read the question properly. Desculpas pela confusão!

 

[berkeman]

So the correct working is:

Please remember that the student must do the bulk of the work on the problem. They should be the one writing out those equations...

 

[Steve4Physics]

Please remember that the student must do the bulk of the work on the problem. They should be the one writing out those equations...

Agreed, thanks. I need to be more sef-discipined.