Kinematics: Emerging From a Tunnel

[deaninator]

Homework Statement

Suppose that a woman driving a Mercedes zooms out of a darkened tunnel at 41.5 m/s. She is momentarily blinded by the sunshine. When she recovers, she sees that she is fast overtaking a truck ahead in her lane moving at the slower speed of 12.9 m/s. She hits the brakes as fast as she can (her reaction time is 0.32 s). If she can decelerate at 2.3 m/s2, what is the minimum distance between the driver and the truck when she first sees it so that they do not collide?

Homework Equations

V=Vo + at
X=1/2(Vo +V)T
X=VoT + 1/2at2
V2=Vo2 + 2ax

(If a 2 is after a letter then it means squared)
X=displacement
A=Acceleration
T=Tome
Vo=Initial Velocity
V=Velocity

The Attempt at a Solution

First I think you need to find the time.which is 41.5/2.3 = 18.04 seconds. Then you need to add on her reaction time (.32 sec) to get 18.36 seconds. Then I used the "X=1/2(Vo +V)T" to get 380.92 meters. I'm not sure if that is correct though. What do you guys think?
Thanks!

 

[brushman]

That doesn't look right to me.

You need to find how much time it takes her to decelerate to the speed of the truck. It is at this point where the distance between the car and the truck stops decreasing.

Then using the time, you can find how much distance is covered. Don't forget to add on extra distance from reaction time.

Also remember you must add the extra distance separately, because until she reacts she is still moving at a constant velocity $v_0$.

 

[deaninator]

That doesn't look right to me.

You need to find how much time it takes her to decelerate to the speed of the truck. It is at this point where the distance between the car and the truck stops decreasing.

Then using the time, you can find how much distance is covered. Don't forget to add on extra distance from reaction time.

"How far will the car travel? How long will that take? How far will the other vehicle travel in that time?" was the hint...

I ended up with 255.978...I found the distance traveled and the time (plus reaction time) of the first car. The I found how far the second vehicle travels in the same time. After that I subtracted the distance traveled by the first car by the distance traveled by the second car.

 

[brushman]

I didn't check your calculations but that should give you the correct answer. 256 seems about right.

 

[deaninator]

I didn't check your calculations but that should give you the correct answer. 256 seems about right.

No it's not. I can't seem to figure it out from here...

 

[brushman]

Okay, I got a different number then you. Keep in mind:

(1) The truck is moving at a constant velocity for the entire time t + reaction time.
(2) You should use the equation $x = v_ot + (1/2)at^2$ for finding the distance traveled by each vehicle.
(3) To find t you use $v = v_0 + at$
(4) Make sure your 'a' is negative, because you are slowing down.

 

[deaninator]

Okay, I got a different number then you. Keep in mind:

(1) The truck is moving at a constant velocity for the entire time t + reaction time.
(2) You should use the equation $x = v_ot + (1/2)at^2$ for finding the distance traveled by each vehicle.
(3) To find t you use $v_o = v + at$
(4) Make sure your 'a' is negative, because you are slowing down.

Ok for the distance of the mercedes i got X=41.5 + (.5)(-2.3)(18.32^2) = -346.15
That can't be right...and what is the acceleration for the truck?

 

[brushman]

First of all, your time is incorrect. Carefully plug your numbers into the equation I wrote above and solve for t. Remember your initial velocity is 41.5, and your final velocity is 12.9. Your acceleration is -2.3.

Secondly, you forgot to multiply 41.5 by the time in your distance equation.

The truck is moving at a constant velocity 12.9. In other words, it has zero acceleration.

 

[deaninator]

First of all, your time is incorrect. Carefully plug your numbers into the equation I wrote above and solve for t. Remember your initial velocity is 41.5, and your final velocity is 12.9. Your acceleration is -2.3.

Secondly, you forgot to multiply 41.5 by the time in your distance equation.

The truck is moving at a constant velocity 12.9. In other words, it has zero acceleration.

Ok so for the car...
X= (41.5)(18.36) + (.5)(-2.3)(18.36^2) = 374.29 m
For the truck...
X= (12.9)(18.36) + (.5) (0) (18.36^2) = 236.84 m

 

[brushman]

Your time is still wrong. Re-calculate it like I said above.

Look at equation (2) above. You must include all variables in the equation.

 

[deaninator]

Your time is still wrong. Re-calculate it like I said above.

Look at equation (2) above. You must include all variables in the equation.

Ok, so right now i have 41.5 = 12.9 + (-2.3)t. T=-12.11 sec...

I think that's wrong. This is a little off topic, but please pardon my ignorance when it comes to physics...I'm just trying to be more proactive nowadays and finishing my homework early on...(my physics teachers constantly pounds 3 to 4 pages of hmwk a day).

 

[brushman]

It's no problem. Being proactive on your homework is a very good idea. The physics forums are a good resource and you should fully take advantage of it (although it shouldn't replace getting direct help from your teacher).

Second, I apologize because I gave you the equation wrong. Luckily, it's really easy to fix and doesn't change much. The real equation is $$v = v_0 + at$$. The only difference is the v and $v_0$ are switched. This will now give you a positive value for time.

Now when you plug in your numbers you will get t = 12.43 s.

 

[brushman]

I'm going to bed now so I will leave you with my final answer which I believe to be correct. I got 187.103 m, and with 2 significant figures that is 190 m. See if you can get it.

Good luck.

 

[deaninator]

It's no problem. Being proactive on your homework is a very good idea. The physics forums are a good resource and you should fully take advantage of it (although it shouldn't replace getting direct help from your teacher).

Second, I apologize because I gave you the equation wrong. Luckily, it's really easy to fix and doesn't change much. The real equation is $$v = v_0 + at$$. The only difference is the v and $v_0$ are switched. This will now give you a positive value for time.

Now when you plug in your numbers you will get t = 12.43 s.

Thank you. Ok so now that we have the time, we need to find the distance of the 2 cars correct?
For the mercedes...
X=(41.5)(12.43) + (.5)(-2.3)(12.43^2). = 338.165 m

For the truck
X=(12.9)(12.43) + (.5)(0)(12.43^2) = 160.347 m

 

[deaninator]

Thank you. Ok so now that we have the time, we need to find the distance of the 2 cars correct?
For the mercedes...
X=(41.5)(12.43) + (.5)(-2.3)(12.43^2). = 338.165 m

For the truck
X=(12.9)(12.43) + (.5)(0)(12.43^2) = 160.347 m

And then do I just subtract those 2 values to get 177.818?

 

[brushman]

You're almost there. The only thing you are forgetting is the reaction time. Thus, the mercedes will go an additional distance of (41.5)(.32), and the truck will go an additional distance of (12.9)(.32).

 

[venkatg]

distance traveled by car = distance traveled by truck + gap ----- (1)

we need to find the gap (the gap when she started pressing her brakes)

let a= deceleration
t = time for the car to come to a stop.
u = initial car speed
v = truck speed
x= gap

From (1) ,

(1/2) a*t*t = v*t + x

we know final car speed is zero.

so 0 = u - a* t and so t = u/a. put this in (1)

x = (u/a)[u/2 - v] = t(u/2 -v)

putting the values, we have t = u/a = 41.5/2.3 = 18.04 secs

So x = 18.04 * (41.5 * 0.5 - 12.9) = 141.64 metres

Accounting her reaction time of 0.32 secs, we have to add the difference of (0.32 * (41.5-12.9)) = 9.15 metres.

So total distance x = 141.64 + 9.15 = 150.8 metres

 

[venkatg]

You're almost there. The only thing you are forgetting is the reaction time. Thus, the mercedes will go an additional distance of (41.5)(.32), and the truck will go an additional distance of (12.9)(.32).

distance traveled by car = distance traveled by truck + gap ----- (1)

we need to find the gap (the gap when she started pressing her brakes)

let a= deceleration
t = time for the car to come to a stop.
u = initial car speed
v = truck speed
x= gap

From (1) ,

(1/2) a*t*t = v*t + x

we know final car speed is zero.

so 0 = u - a* t and so t = u/a. put this in (1)

x = (u/a)[u/2 - v] = t(u/2 -v)

putting the values, we have t = u/a = 41.5/2.3 = 18.04 secs

So x = 18.04 * (41.5 * 0.5 - 12.9) = 141.64 metres

Accounting her reaction time of 0.32 secs, we have to add the difference of (0.32 * (41.5-12.9)) = 9.15 metres.

So total distance x = 141.64 + 9.15 = 150.8 metres

 

[venkatg]

distance traveled by car = distance traveled by truck + gap ----- (1)

we need to find the gap (the gap when she started pressing her brakes)

let a= deceleration
t = time for the car to come to a stop.
u = initial car speed
v = truck speed
x= gap

From (1) ,

(1/2) a*t*t = v*t + x

we know final car speed is zero.

so 0 = u - a* t and so t = u/a. put this in (1)

x = (u/a)[u/2 - v] = t(u/2 -v)

putting the values, we have t = u/a = 41.5/2.3 = 18.04 secs

So x = 18.04 * (41.5 * 0.5 - 12.9) = 141.64 metres

Accounting her reaction time of 0.32 secs, we have to add the difference of (0.32 * (41.5-12.9)) = 9.15 metres.

So total distance x = 141.64 + 9.15 = 150.8 metres

 

[brushman]

The final speed of the mercedes is not zero. The car only needs to slow down to 12.9 m/s, the speed of the truck, to avoid running into it.

 

[venkatg]

####################################$# # MISTAKE CORRECTION # !Please ignore earlier and read this!$#####################################

Let us take all distances and speed with respect to an observer on the ground.

distance traveled by car = distance traveled by truck + gap ----- (1)

we need to find the gap (the gap when she started pressing her brakes)

let a= deceleration
t = time for the car to come to a stop.
u = initial car speed
v = truck speed
x= gap

From (1) and using the formula v^2 = u^2 - 2*a*s (s = distance traveled by car during the braking effort),

(u^2 - v^2)/(2*a) = v*t + x ----------------------- (2)

Also we know final car speed (with respect to the ground) is 12.9m/sec (truck speed)

so v = u - a* t and so t = (u-v)/a. put this in (2)

x = (u-v) * (u-v)/(2*a)

putting the values, we have time

So x = 178 metres

Accounting her reaction time of 0.32 secs, we have to add the difference of (0.32 * (41.5-12.9)) = 9.15 metres.

So total distance x = 178 + 9 = 187 metres

Time to brake the car to the truck speed is (u -v)/a = 12.43 secs

 

[venkatg]

And then do I just subtract those 2 values to get 177.818?

Yes that's right. Also add the distance covered during the reaction time

 

[zgozvrm]

Now that everyone seems to have agreed on the answer, I'd like to share my thoughts on solving this. It's really the same thing that venkatg did, just described more clearly (I think):


The car's initial velocity is [itex]V_{ic} = 41.5 m/s[/tex]
The truck's velocity is constant [itex]V_T = 12.9 m/s[/tex]
The car's acceleration rate is [itex]a = -2.3 m/s^2[/tex]
The car driver's reaction time is [itex]t_{react}=0.32 s[/tex]

The car must slow down to the the truck's speed before hitting it, so the car's final velocity is [itex]V_{fc} = V_T = 12.9 m/s[/tex]

The equation for final velocity given initial velocity, acceleration and time is
[itex]V_f = V_i + at[/tex]

We know all but "t" for the car, so [itex]41.5 = 12.9 - 2.3t_{brake}[/tex]
Which becomes [itex]t_{brake}=(12.9 - 41.5)/2.3 \approx 12.43 s[/tex]

Then we have for the car:
Distance traveled during reaction time [itex]D_{react} = V_{ic} \times t_{react} = 41.5 \times 0.32 = 13.28 m[/tex]
Distance traveled during braking
$$D_{brake} = V_{ic} \times t_{brake} + \frac{a}{2} (t_{brake})^2 \approx 41.5 \times 12.43 - \frac{2.3}{2} \times 12.43^2 \approx 338.23 m$$

The car's total displacement is then [itex]D_{car} = D_{react} + D_{brake} \approx 13.28 + 338.23 \approx 351.51[/tex]


During this time ([itex]t_{tot} = t_{react} + t_{brake} \approx 0.32 + 12.43 \approx 12.75 s[/tex]), the truck travels
[itex]D_{truck} = V_T \times t_{tot} \approx 12.9 \times 12.75 \approx 164.54 m[/tex]
after which the Mercedes would be right on the truck's tail, so...

The minimum distance between them when the Mercedes driver first sees the truck is then given as [itex]D_{min} \approx (D_{car} - D_{truck}) \approx 351.51 - 164.54 \approx 186.97 m[/tex]



Note: Calculations were made with actual values, not rounded (displayed) values.

 

[venkatg]

Looks very clear zgozvrm!