# Velocity, acceleration, collision

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1. Oct 14, 2015

### PhysicsBoyMan

1. The problem statement, all variables and given/known data
Speedy Sue, driving at 30m/s, enters a one way tunnel. She observes a slo-moving van 155m ahead traveling at 5m/s. Sue applies her brake and accelerates at 2m/s. Will there be a collision? If yes, determine how far into the tunnel and at what time. If no, determine the distance of closest approach between Sue's car and the van.

2. Relevant equations
V1 = 30m/s
V2 = 5m/s
x = 155m
a= -2m/s2

3. The attempt at a solution
I attempted to make Δxsue = Δxvan.
I used Δx = Vot + 1/2 at2.
I combined like terms and tried to use the quadratic equation. I had a negative number under my square root.
I am wondering if this is the right approach.

I used the same formula for Sue only, and found that it would take 6.63s to move the 155m with her velocity and acceleration.

I've been at this for a long time and have tried many different ways. I do have a list of equations for when acceleration is constant. A little bit of guidance here would help a lot, I have spent hours now and an outside perspective would be refreshing.

2. Oct 14, 2015

### collinsmark

Yes, that's the correct approach.
That's one equation. The one for Speedy Sue. What's the other equation?

'Sounds like you are on the right track. A quadratic equation is involved.

Something must have went wrong around there then. The value under my square root was positive.

Please show your work thus far and maybe we can point you in the right direction.
That won't help you a whole lot. By the time Sue is 155 meters into the tunnel, the Van isn't at that location anymore. It moved.

Try going back to solving for t and we'll see where that takes us.

3. Oct 14, 2015

### collinsmark

By the way, just to be clear, there isn't anything fundamentally wrong with getting negative numbers under the square root for problems like these. It would just mean that a collision won't happen. It means that there is no such (real) t that exists that would cause the displacements to be equal.

However, in this particular problem, the sign under the square root is positive, according to my calculations. So you'll have to show your work before we can continue.

4. Oct 14, 2015

### PhysicsBoyMan

That's a great point Collinsmark! Thanks for teaching that.

This is how far I got. To get a negative for 4ac either a or c are wrong. A is -1 it seems straight forward I think my c is wrong. The way I learned to do delta x is x final - x initial so I'm not sure where I went wrong but that must be the problem I would think.

Thanks for the help.

screen capture

5. Oct 14, 2015

### Mister T

It looks like you've got an initial position $x_o$ of 155 m for the van, which is correct, but it looks like you've set $\Delta x$ equal to 155 m for Sue.

You should be setting $x_{Sue}=x_{van}$,

but it looks like instead you're setting $x_{van}=0$.

$\Delta x=v_ot+\frac{1}{2}at^2$

try using

$x=x_o+v_ot+\frac{1}{2}at^2$

for both Sue and the van.

Last edited: Oct 14, 2015
6. Oct 14, 2015

### collinsmark

Hmm.

Let's step back a bit.

Equation for Sue:

$x = v_0 t + \frac{1}{2}a t^2$

You had that, however, don't put in 155 m for $x$ in that equation. We don't know what $x$ is. At least we don't know what it is yet. It's not 155 m though. It's farther than that. It's the distance that Speedy Sue will travel before reaching the van. But remember the van is moving, so that distance is something greater than 155 meters.

But again, don't worry about its actual value yet.

Equation for the van:

That one needs work. Try that equation again. Recall the van is not accelerating. It's simply moving at a constant velocity. What's the relationship between displacement and time of something moving at a constant velocity?

For this equation you'll want to specify the position of the van relative to the entrance to the tunnel (That way Sue's displacement and the van's displacement are both relative to the same location). And it's this equation where the 155 m fits in.

7. Oct 14, 2015

### Mister T

The equation for $x_{van}$ is almost correct.

8. Oct 14, 2015

### PhysicsBoyMan

I forgot to take a picture of the complete solution, but I finished it thanks to you guys. That one took a long time!

I used x=v0t+12at2 for sue and x=vt for van. Sue's x position was unknown as you say, as opposed to 150m. The time ended up being 11.something seconds.

I will be more careful in the future and think before I declare something to be true in my equations.

Thanks again!

9. Oct 15, 2015

### Sbr:132

There will be No collision.and the distance between them will be 5 m.

10. Oct 15, 2015

### Sbr:132

Firtly we need to find how much time sue's car take to stop
V=u+at
0=30+(-2*t)
-30=-2t

t=15
in that 15 sec how much the slow van will go
s=vt=15*5=75
and after acclerating at -2ms-2 sues car will stop
after 15 s and the distance the car will reach
S=ut+1/2at2=30*15+1/2*(-2)*(15)2=225m
in that 15 sec how much the slow van will go
s=vt=15*5=75
and after 15 s
the van will be from beginig of car's position (155+75)=230m
so there will be no collision cause the car goes less distance than 230 m as the car goes 225m
and the closet distance between them is (230-225)=5 m
do it like that

11. Oct 15, 2015

### haruspex

This approach will not work. Indeed, you got the wrong answer. They collide before Sue comes to a stop.

12. Oct 15, 2015

### Sbr:132

[Q..UOTE="haruspex, post: 5258893, member: 334404"]This approach will not work. Indeed, you got the wrong answer. They collide before Sue comes to a stop.[/QUOTE]
Ok.
plz do the right answer for me.

13. Oct 15, 2015

### haruspex

Consider the problem in the reference frame of the lead vehicle. How far does Sue start from it and how is Sue moving in relation to it? At their closest point, what will the relative velocity be?

14. Oct 15, 2015

### Mister T

You calculate 15 s for Sue to come to a stop. The OP calculated approximately 11 s before a collision, so if he's right and you're right, Sue will hit the van before she can come to a stop!

We're supposed to be helping the OP, not solving the problem.

Hint: The position-time graph of Sue's motion is a concave-down parabola with the vertex at t = 15 s. The van's position-time graph is a straight line with a slope of 5 m/s. The line crosses the parabola in two places,

15. Oct 15, 2015

### Mister T

Use

$x=x_o+v_ot+\frac{1}{2}at^2$

for both Sue and the van. Then set the two values of $x$ equal to each other and solve for $t$. You will get two times, one for each intersection of the parabola with the line. One before the vertex, the other after.

16. Oct 15, 2015

### Sbr:132

Firtly we need to find how much time sue's car take to stop
V=u+at
0=30+(-2*t)
-30=-2t
t=15
and after acclerating at -2ms-2 sues car will stop after 15 s and the distance the car will reach
S=ut+1/2at2=30*15+1/2*(-2)*(15)2=225m
in that 15 sec how much the slow van will go
s=vt=15*5=75
and after 15 s the van will be from beginig of car's position (155+75)=230m
so there will be no collision cause the car goes less distance than 230 m as the car goes 225m
and the closet distance between them is (230-225)=5 m
if u have any solution plz do that sum.
Now I really want to understand it.
wather he is right or me.

17. Oct 15, 2015

### haruspex

we have been around this already. That is your assumption, but it is not right. In the final moments before Sue stops, Sue will be moving more slowly than the van, so they will be getting further apart. Therefore the closest approach is before Sue stops.

18. Oct 15, 2015

### Mister T

$x_{Sue}=(0)+(30)t+\frac{1}{2}(-2)t^2$,

$x_{van}=(155)+(5)t+\frac{1}{2}(0)t^2$.

19. Oct 15, 2015

### Sbr:132

look you almost 40 yrs elder than me.
if my sum below is worng do the right sum for me mathematically and prove that there will be a collision.I'm just a college student.and I want to understand it.
my sum is below
Firtly we need to find how much time sue's car take to stop
V=u+at
0=30+(-2*t)
-30=-2t
t=15
again
after acclerating at -2ms-2 sues car will stop after 15 s and the distance the car will reach
S=ut+1/2at2=30*15+1/2*(-2)*(15)2=225m
again
in that 15 sec how much the slow van will go
s=vt=15*5=75
and after 15 s the van will be from beginig of car's position (155+75)=230m
so there will be no collision cause the car goes less distance than 230 m as the car goes 225m
and the distance between them is (230-225)=5 m

Last edited: Oct 15, 2015
20. Oct 15, 2015

### Mister T

Follow the instructions I gave in Post #18. If you have specific questions about that procedure ask and we will help. But we won't do the work for you.

(The basic flaw in your method is that you calculate the time it takes Sue to stop. Instead the solution of the problem involves calculating the time it takes Sue to collide. Stopping is the not necessarily the same thing as colliding. We will not attempt to prove that we are right, but we will try to guide you towards proving it to yourself. We won't do the work for you. We will help you do the work for yourself.)

Last edited: Oct 15, 2015