How far apart are the stop signs?

[Abdul.119]

Homework Statement

A car starts from rest at a stop sign. It accelerates at 3.7m/s^2 for 7.3s , coasts for 1.6s , and then slows down at a rate of 3.3m/s^2 for the next stop sign.
How far apart are the stop signs?

Homework Equations

d = vt +1/2 a*t^2
Vf = Vi + a*t
Vf^2 = Vi^2 + 2a*s

The Attempt at a Solution

I believe I should calculate the distance over the first 7.3 seconds, which would be 1/2*3.7*(7.3^2) = 98.6 m.. ok when it coasts for 1.6s, I think we must assume there is no friction? so the velocity from there is Vf^2 = 0 + 2*3.7*98.6, Vf = 27m/s, traveling at that speed for 1.6s would travel 43.216 meters.. now I'm not sure how to find the distance traveled in the last part, but I know it took the car 27/3.3 = 8.1 seconds to decelerate from 27 m/s to 0 m/s .. how do I find that last distance? I tried the first equation which gave me 110.5 meters.. is that correct? and the total distance from the two stop signs would be 252 m? (I don't want to put a wrong answer because it gets penalty)
Thanks

 

[collinsmark]

Hello Abdul.119,

Welcome to Physics Forums!

Homework Statement

A car starts from rest at a stop sign. It accelerates at 3.7m/s^2 for 7.3s , coasts for 1.6s , and then slows down at a rate of 3.3m/s^2 for the next stop sign.
How far apart are the stop signs?

Homework Equations

d = vt +1/2 a*t^2
Vf = Vi + a*t
Vf^2 = Vi^2 + 2a*s

The Attempt at a Solution

I believe I should calculate the distance over the first 7.3 seconds, which would be 1/2*3.7*(7.3^2) = 98.6 m..

So far so good.

ok when it coasts for 1.6s, I think we must assume there is no friction?

I think that is a valid assumption. The problem didn't mentioned anything about friction, so I would just ignore friction.

so the velocity from there is Vf^2 = 0 + 2*3.7*98.6, Vf = 27m/s, traveling at that speed for 1.6s would travel 43.216 meters..

[STRIKE]You'll have to rethink that.

During that stretch, the car is coasting (with no friction). That means the final velocity (for this stretch) is equal to the initial velocity (again though, just for this stretch). What does that tell you about the car's acceleration (for this stretch)?[/STRIKE]

Edit: Oh, I see what you did. You're calculating the final velocity after the first stretch (which is also the final velocity after the second stretch). I would have used the v = at formula, which gives you the same answer. But your approach also works.

now I'm not sure how to find the distance traveled in the last part, but I know it took the car 27/3.3 = 8.1 seconds to decelerate from 27 m/s to 0 m/s .. how do I find that last distance? I tried the first equation which gave me 110.5 meters.. is that correct? and the total distance from the two stop signs would be 252 m? (I don't want to put a wrong answer because it gets penalty)
Thanks

Think about the conditions in the final stretch. The car is slowing down instead of speeding up, but it does have a nonzero, uniform acceleration. The car's initial velocity in the final stretch is not zero. The car's final velocity is zero. You don't know how long it takes the car on the final stretch, but you don't care either. Can you find a kinematics equation that goes with this?

Edit: Finding the time it takes to complete the final stretch, and then plugging that into the first equation which you used is another way to get the distance of the final stretch. But my point above is that there is a kinematics formula that will allow you to do this all in one step.

 

[mfb]

Friction is not relevant, only accelerations are interesting here.

You'll have to rethink that.

During that stretch, the car is coasting (with no friction). That means the final velocity (for this stretch) is equal to the initial velocity (again though, just for this stretch). What does that tell you about the car's acceleration?

That part is still fine. It is a complicated way to calculate the velocity, but it is possible.

now I'm not sure how to find the distance traveled in the last part, but I know it took the car 27/3.3 = 8.1 seconds to decelerate from 27 m/s to 0 m/s .. how do I find that last distance? I tried the first equation which gave me 110.5 meters.. is that correct? and the total distance from the two stop signs would be 252 m? (I don't want to put a wrong answer because it gets penalty)

Yes it is correct.

I'm not sure how sensitive the system is in terms of rounding errors. In doubt, use more digits in the intermediate steps.

 

[collinsmark]

That part is still fine. It is a complicated way to calculate the velocity, but it is possible.

Correct. I've modified my reply above to reflect that. (And I've reminded myself again to finish coffee before posting.)

Abdul.119,
As you've discovered in this case, you'll often find that there is more than one way to solve such physics problems. You can often save yourself some effort though by using a kinematics formula that best matches the situation at hand. In general, examine the problem to find which variables you know, which variables you don't know but would like to, and which variables you don't care about. Then choose the kinematics formula that doesn't contain variables that you don't care about, but allows you to solve for those which you need. (And of course, make sure that formula is applicable to the situation, such as constant velocity, uniform acceleration restrictions, etc.) There's nothing inherently wrong with using different, multiple formulas, and solving for the intermediate variables, but choosing the right formula at the beginning will save you time when taking exams.

 

[HalfLostAndSearching]

for your question! I would approach this problem by first defining the variables and equations that are relevant to the situation. The distance between the two stop signs can be represented by the variable "d". The acceleration of the car can be represented by "a" and the time it takes to accelerate and decelerate can be represented by "t". The initial velocity of the car is 0 m/s and the final velocity is also 0 m/s.

Using the equation d = vt + 1/2at^2, we can calculate the distance traveled during the acceleration phase:
d1 = (0)(7.3) + 1/2(3.7)(7.3^2) = 98.6 m

During the coasting phase, the car maintains a constant velocity of 27 m/s for 1.6 seconds. Therefore, the distance traveled during this phase is:
d2 = (27)(1.6) = 43.2 m

Lastly, during the deceleration phase, we can use the equation Vf^2 = Vi^2 + 2ad to find the distance traveled:
0^2 = (27)^2 + 2(-3.3)d
d = 110.5 m

Therefore, the total distance traveled between the two stop signs is:
d = d1 + d2 + d3 = 98.6 m + 43.2 m + 110.5 m = 252.3 m

Based on the given information, it seems that your calculations and answer are correct. However, as a scientist, I would recommend double-checking your work and making sure all units are consistent. Additionally, it is always helpful to include units in your final answer.