Constant acceleration problem

[HRubss]

Homework Statement

A car, length 2.4m, traveling at 26m/s wants to pass a large truck, length 12m, also traveling at 26m/s. If the car can accelerate at 4.4m/s², how much time will it take the car to pass the truck (note the back of the car has to be in front of the truck)? How far will the car have traveled during this time? How much time did it take the car to pass the truck?

Homework Equations

v₁² = v₀² +2a(x₁-x₀)
v₁² = v₀² + a(t)

The Attempt at a Solution

I started off with plugging in the numbers in the first equation to get v₁ and then plugged that in the second equation to get t.
v₁² = (24m/s)² + 2(4.4m/s²)(12m-2.4m)
(≈26m/s)=(24m/s)+(4.4m/s²)(t) which gave me t ≈ 0.39s
but I feel like that only gives me the time it takes to speed up to the truck and not actually pass it. So then I arbitrarily used 13m and inserted that in the first equation as x₁ instead of 12m since 13m is further than 12m but that still gave me ≈26m/s which is the same velocity that the truck is driving so I wouldn't think it would pass the truck either, t ≈ 0.42s in this case.
I appreciate any help!

 

[haruspex]

12m-2.4m

What is the reasoning behind that operation?

 

[HRubss]

What is the reasoning behind that operation?

If the car is 2.4m and the truck is 12m, the car would have to travel 9.4m to be at the same "point" of the truck and that's when I realized that it didn't seem right which is why I used 13m instead.

 

[NFuller]

You should try using the formula
$$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$$
It would also be simpler to use speeds and distances relative to the truck.

but I feel like that only gives me the time it takes to speed up to the truck and not actually pass it. So then I arbitrarily used 13m and inserted that in the first equation as x1 instead of 12m since 13m is further than 12m

So if the car has to first catch up to the 12m truck and then needs to travel another 2.4m to get its bumper in front of the truck, what distance must the car travel?

 

[HRubss]

You should try using the formula
$$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$$
It would also be simpler to use speeds and distances relative to the truck.

So if the car has to first catch up to the 12m truck and then needs to travel another 2.4m to get its bumper in front of the truck, what distance must the car travel?

Did not cross my mind to think of it in that way, so the car would have to travel 14.4m to get his bumper in front of the truck.
Using that formula provided and plugging in the numbers
14.4m = 2.4m + (26m/s)(t) + ½(4.4m/s²)t², subtracting 2.4m from 14.4m and then bring the 12m to the other side again to get a quadratic equation and then using the quadratic formula, I got t ≈ 0.44s?

 

[haruspex]

If the car is 2.4m and the truck is 12m, the car would have to travel 9.4m to be at the same "point" of the truck

So if, instead, the truck were passing the car it would have to go 2.4m-12m?

 

[haruspex]

14.4m = 2.4m +

You just, correctly, added the 2.4m to the 12m, but by writing it also on the other side of the equation you take it off again.

+ (26m/s)(t)

Has the truck stopped?

 

[HRubss]

You just, correctly, added the 2.4m to the 12m, but by writing it also on the other side of the equation you take it off again.

Has the truck stopped?

But the car want's to reach 14.4m and that would be his x_final and the Δx is 12m? The cars velocity and trucks velocity are both 26m/s? I apologize if I'm being dense, I'm use to question that we've done in class and this wasn't one of them.

 

[NFuller]

$\Delta x$ is 14.4m since that is the distance the car must travel to get in front of the truck.

The other issue is that you need to be working in the reference frame of the truck not the road. You inserted the velocity as 26m/s which means you are measuring the displacement with respect to the road. This is obviously larger than 14.4m since as the car tries to move this distance in front of the truck, the vehicles will have both traveled a much larger distance down the road.

 

[haruspex]

But the car want's to reach 14.4m and that would be his xfinal and the Δx is 12m?

Concentrate on one part of the car, the back end say. Relative to the truck, how far does the back end move?

The cars velocity and trucks velocity are both 26m/s?

Consider what equation you would have written if the truck had been stationary. It would have been the equation you wrote in post #5, right?
It must take longer to pass a moving truck.

As NFuller suggests the easiest way to solve thos problem is to work entirely in the reference frame of the truck. From that perspective, what is the distance the car moves (14.4m, as found already) and what is its initial speed?

If you are not confident working in different reference frames, you need an equation for where the truck will be at time t. The car needs to go 14.4m further in the same time.

 

[HRubss]

As NFuller suggests the easiest way to solve thos problem is to work entirely in the reference frame of the truck. From that perspective, what is the distance the car moves (14.4m, as found already) and what is its initial speed?

Well if the truck were stationary, it would take 0.45s to get the car's rear bumper 14.4m or in front of the truck. Using that time, the truck while moving, at 0.45s, will have traveled another 11.7m? 11.7+14.4m= 26.1m and I use that as the x distance that the car needs to travel all together. After using equation in post 5 and again applying the quadratic formula, I got 0.93s which doesn't seem right to me since that would mean the car would be accelerating pretty fast to reach 26m..

 

[NFuller]

Well if the truck were stationary, it would take 0.45s to get the car's rear bumper 14.4m or in front of the truck.

How did you get this number? This is the problem you need to solve, everything else you did was unnecessary.

 

[HRubss]

 

[haruspex]

will have traveled another 11.7m? 11.7+14.4m= 26.1m and I use that as the x distance that the car needs to

But in the extra time (.93s-.45s) the truck will have moved on further.
It is going to take a long time to get to the answer working that way.
Suppose it takes t secounds to pass the truck. How far will the truck have moved? What is the total diistance the car has to move in the same time?

 

[HRubss]

How far will the truck have moved? What is the total diistance the car has to move in the same time?

This I'm not sure of

 

[NFuller]

Again if we work in the frame of the truck, the problem is fairly simple. At time zero, the car and truck are moving at the same speed, so if you are in the truck looking at the car, at what speed does the car appear to be moving?

 

[HRubss]

Again if we work in the frame of the truck, the problem is fairly simple. At time zero, the car and truck are moving at the same speed, so if you are in the truck looking at the car, at what speed does the car appear to be moving?

Well if they're both moving at the same speed, it would look like the car isn't moving. While the car's accelerating, then.. (currently out atm!)

?

 

[NFuller]

Well if they're both moving at the same speed, it would look like the car isn't moving

Right so $v_{0}=0$ in the one dimensional motion equation
$$\Delta x =v_{0}+\frac{1}{2}at^{2}$$
You already know $\Delta x$ and $a$ so what is $t$?

 

[HRubss]

Right so $v_{0}=0$ in the one dimensional motion equation
$$\Delta x =v_{0}+\frac{1}{2}at^{2}$$
You already know $\Delta x$ and $a$ so what is $t$?

2.52s?

 

[NFuller]

Did you set $\Delta x=14.4$m?

Your answer is slightly off.

 

[HRubss]

Did you set $\Delta x=14.4$m?

Your answer is slightly off.

Whoops, divided by 14m instead. So the answer was t=2.55s? That's the same thing I got when I did it on the napkin.

 

[NFuller]

Yes, but this was much faster!

 

[HRubss]

Yes, but this was much faster!

Oh wow, that would have required a lot of critical thinking out of me. It seems so simple now.. thanks for the help guys!

 

[haruspex]

Oh wow, that would have required a lot of critical thinking out of me. It seems so simple now.. thanks for the help guys!

I'd like to add a couple of observations.

Ok, that was using relative motion, i.e. the reference frame of the truck, and that is definitely the easier way. Just to see it can also be done in the ground frame:
Say at time 0 the car is immediately behind the truck and the rear of the car is position zero.
At time t the rear of the car is at the front of the truck. Writing that as an equation:
2.4+12+26t = 0+26t+½4.4t².
The cancellation of the two 26t is equivalent to working in the truck's frame.

Secondly, I feel the need to point out that the question gives a totally unrealistic impression of the room needed for overtaking. We do not drive bumper to bumper before pulling out. In England, the official recommendation was, and maybe still is, the two second rule, on which basis the car would have needed to start moving to the overtaking lane when closer than 2x26m. Typically, drivers obey more like a one second rule.
So, more realistically, but slightly simpler numbers...
Start 25m behind at 26m/s, accelerate at 4.5m/s², car+truck length = 14m, finish 25m in front:
25+14+25=½ 4.5t²
t=5.333s.
Top speed= 26+4.5 t = 50m/s = 180 kph.
So now we have another problem, going way over any reasonable speed limit. A limit of 110kph is about 30.5 m/s. The car would reach that in one second, so nearly all the passing is done at the limit. That pushes the time beyond 64/(30.5-26)=14.2s, and the distance in the passing lane to over 430m.

 

[HRubss]

I'd like to add a couple of observations.

Ok, that was using relative motion, i.e. the reference frame of the truck, and that is definitely the easier way. Just to see it can also be done in the ground frame:
Say at time 0 the car is immediately behind the truck and the rear of the car is position zero.
At time t the rear of the car is at the front of the truck. Writing that as an equation:
2.4+12+26t = 0+26t+½4.4t².
The cancellation of the two 26t is equivalent to working in the truck's frame.

I know how a relativity is important but my professor never taught us how to use it in any class demonstration nor has he told us to create some sort of equation. I have another 2-D projectile motion problem and also a free-fall problem that I don't understand either. Should I make another thread for them or just add them here? These problems are off of a study guide for a test I have on Wednesday.

 

[haruspex]

I know how a relativity is important but my professor never taught us how to use it in any class demonstration nor has he told us to create some sort of equation. I have another 2-D projectile motion problem and also a free-fall problem that I don't understand either. Should I make another thread for them or just add them here? These problems are off of a study guide for a test I have on Wednesday.

Please start a new thread for each problem.