Kinematic problem, are my variables right,

[rashad764]

Homework Statement

A car starts from rest at a stop sign. It accelerates at 4.3 m/s2 for 6.0 s , coasts for 2.5 s , and then slows down at a rate of 2.5 m/s2 for the next stop sign. How far are the stops signs apart.
x0=0m
v0=0m/s
t0=0s
I put 0 because it says a car stars from rest. Not sure if this is right
x1=?
v1=?=25.8m/s
t1=6.0s

x2=?
v2=? 10.75 m/s
t2=2.5s

x3=?
v3=?
t3=?[/B]

Homework Equations

trying to use the constant acceleration model[/B]
x3=x2+v2(t3-t12)+1/2a(t3-t2)

The Attempt at a Solution

i need to find the velocity first. I'm using v1=v0+a(t1-t0)=0+4.3(6)=25.8m/s
I then velocity for x2. v2=v1+a(t2-t1)= 25.8+4.3(2.5-6)=10.75 m/s
I want to if everything up to know is done right.
After this i plan to use this formula to find x2 and x1

x1=x0+v0(t1-t0)
x2=x1+v1(t2-t1)
[/B]

 

[MarkFL]

Let's assume all distances are in meters, and time is measured in seconds. I would use the kinematic formula:

$x(t)=\frac{1}{2}at^2+v_0t+x_0\tag{1}$

Let's orient our coordinate axis such that the car begins at the origin, i.e.:

$x_0=0$

We are told the car begins at rest (what is $v_0$?), and we are told the car accelerates at $4.3\,\frac{\text{m}}{\text{s}^2}$ for 6 seconds...when you plug all these values into (1), what do you have?

 

[rashad764]

Let's assume all distances are in meters, and time is measured in seconds. I would use the kinematic formula:

$x(t)=\frac{1}{2}at^2+v_0t+x_0\tag{1}$

Let's orient our coordinate axis such that the car begins at the origin, i.e.:

$x_0=0$

We are told the car begins at rest (what is $v_0$?), and we are told the car accelerates at $4.3\,\frac{\text{m}}{\text{s}^2}$ for 6 seconds...when you plug all these values into (1), what do you have?

77.4
if the car is at rest is it not 0?

 

[MarkFL]

77.4
if the car is at rest is it not 0?

Yes, since the care is initially at rest, we have $v_0=0$...and so let's check your result:

$x(6)=\frac{1}{2}\cdot4.3\cdot6^2=77.4\quad\checkmark$

Okay, now we need to determine the distance traveled during the second phase, when the car coasts. The initial velocity for this phase is the velocity attained at the end of the first phase, and for this, we can use the kinematic equation:

$v(t)=at+v_0\tag{2}$

Plugging in the values relevant to the first phase, what is the velocity of the car at the end of the first phase?

 

[rashad764]

Yes, since the care is initially at rest, we have $v_0=0$...and so let's check your result:

$x(6)=\frac{1}{2}\cdot4.3\cdot6^2=77.4\quad\checkmark$

Okay, now we need to determine the distance traveled during the second phase, when the car coasts. The initial velocity for this phase is the velocity attained at the end of the first phase, and for this, we can use the kinematic equation:

$v(t)=at+v_0\tag{2}$

Plugging in the values relevant to the first phase, what is the velocity of the car at the end of the first phase?

25.8?

 

[MarkFL]

25.8?

Yes, that car's velocity at the end of the first phase, which will be the initial velocity of the second phase. What are the initial position and acceleration for the second phase?

 

[rashad764]

Yes, that car's velocity at the end of the first phase, which will be the initial velocity of the second phase. What are the initial position and acceleration for the second phase?

is the acceleration still 4.3 m/s^2?

is the initial position calculated with the first formula you posted but instead of v0 and x0, we use v1 and x1?

 

[MarkFL]

is the acceleration still 4.3 m/s^2?

If the car is coasting, then it's velocity is remaining constant (not changing), so what does this tell us about the acceleration?

is the initial position calculated with the first formula you posted but instead of v0 and x0, we use v1 and x1?

The initial position for the second phase is the final position of the first phase. I would use $v_0$ and $x_0$ for the second phase, just keep in mind that these are independent of the initial values used for the other phases.

 

[rashad764]

If the car is coasting, then it's velocity is remaining constant (not changing), so what does this tell us about the acceleration?
so its 0The initial position for the second phase is the final position of the first phase. I would use $v_0$ and $x_0$ for the second phase, just keep in mind that these are independent of the initial values used for the other phases.

so would it be i would be using 1/2(0)(2.5)+25.8+77.4?

 

[MarkFL]

so would it be i would be using 1/2(0)(2.5)+25.8+77.4?

You are correct about the acceleration for the second phase being 0. So, plugging into (1), we obtain:

$x(2.5)=\frac{1}{2}\cdot0\cdot2.5^2+25.8\cdot2.5+77.4=?$

 

[rashad764]

You are correct about the acceleration for the second phase being 0. So, plugging into (1), we obtain:

$x(2.5)=\frac{1}{2}\cdot0\cdot2.5^2+25.8\cdot2.5+77.4=?$

141.9m

 

[rashad764]

how will i find the time for the last phase?

 

[MarkFL]

141.9m

Yes, that's correct. Now, for the third phase, we are only given the acceleration. We know the initial and final velocities as well as the initial position , so while we could determine the time using (2), let's instead use the following kinematic equation:

$x_f=\dfrac{v_f^2-v_i^2}{2a}+x_i\tag{3}$

Can you first identify all of the values of the parameters on the right side of the equation?

 

[rashad764]

is initial velocity 25.8 and initial position 141.9? how do i figure out the final velocity?

 

[MarkFL]

is initial velocity 25.8 and initial position 141.9? how do i figure out the final velocity?

During the second phase the velocity remains unchanged, so yes the initial velocity for the third and final phase will have this value. You are correct too about the initial position. The final velocity is the car's speed at the end of the third phase...where there is a stop sign, so what will the car do there?

 

[rashad764]

During the second phase the velocity remains unchanged, so yes the initial velocity for the third and final phase will have this value. You are correct too about the initial position. The final velocity is the car's speed at the end of the third phase...where there is a stop sign, so what will the car do there?

it will come to a stop so it is 0

 

[MarkFL]

it will come to a stop so it is 0

Correct, so you now have everything you need to answer the question...what is the total distance traveled between stop signs?

 

[rashad764]

Correct, so you now have everything you need to answer the question...what is the total distance traveled between stop signs?

im getting 8.7s seconds for t3 , is this correct?

 

[MarkFL]

im getting 8.7s seconds for t3 , is this correct?

No, but you don't need to know the time that elapses during the third phase, because we are using an equation that doesn't require time. However, if you wish to compute this time, you can use (2) to do so...how did you arrive at the value of 8.7 s?

 

[rashad764]

No, but you don't need to know the time that elapses during the third phase, because we are using an equation that doesn't require time. However, if you wish to compute this time, you can use (2) to do so...how did you arrive at the value of 8.7 s?

i didn't read the formula correctly. the 3rd formula is for the final distance correct

 

[MarkFL]

i didn't read the formula correctly. the 3rd formula is for the final distance correct

Yes, (3) will give you the total distance traveled for all 3 phases.

 

[rashad764]

Yes, (3) will give you the total distance traveled for all 3 phases.

0^2-28.5^2/2(2.5) +141.9?

 

[MarkFL]

0^2-28.5^2/2(2.5) +141.9?

Let's convert your result to $\LaTeX$...

$x_f=\dfrac{0^2-28.5^2}{2(2.5)} +141.9$

The only issue I see is your acceleration...since the car is slowing down, what should the sign of the acceleration be?

 

[rashad764]

Let's convert your result to $\LaTeX$...

$x_f=\dfrac{0^2-28.5^2}{2(2.5)} +141.9$

The only issue I see is your acceleration...since the car is slowing down, what should the sign of the acceleration be?

negative right
total distance is 275m

 

[rashad764]

my professor gave us these 3 formulas to use v2=v1+a(t2-t1)
x2=x1+v1(t2-t1)+1/2a(t2-t1)
v2^2=v1^2+2a(x2-x1)however the formulas that was being used here seem way more simple and easier to understand

 

[rashad764]

would you mind posting the kinematic formulas that are needed to solve these types of motion problems. You really helped me understand this.

 

[rashad764]

would you mind posting the kinematic formulas that are needed to solve these types of motion problems. You really helped me understand this.

or if you could link where you found them

 

[MarkFL]

my professor gave us these 3 formulas to usev2=v1+a(t2-t1)
x2=x1+v1(t2-t1)+1/2a(t2-t1)
v2^2=v1^2+2a(x2-x1)however the formulas that was being used here seem way more simple and easier to understand

Those 3 formulas are the same as the formulas I gave, just written in slightly different ways.

I was using:

$t=\Delta t=t_2-t_1$

and so the first formula you posted could be written:

$v_2=at+v_1$

I was using $v_2=v(t)$ and $v_1=v_0$ and so this becomes:

$v(t)=at+v_0$

Similarly, the second equation you posted (you should have the time difference square in the last term) becomes:

$x(t)=\frac{1}{2}at^2+v_0t+x_0$

And the third equation you posted can be arranged and written as:

$x_f=\dfrac{v_f^2-v_i^2}{2a}+x_i$

These equations are really all you need for problems involving constant acceleration.

 

[rashad764]

Those 3 formulas are the same as the formulas I gave, just written in slightly different ways.

I was using:

$t=\Delta t=t_2-t_1$

and so the first formula you posted could be written:

$v_2=at+v_1$

I was using $v_2=v(t)$ and $v_1=v_0$ and so this becomes:

$v(t)=at+v_0$

Similarly, the second equation you posted (you should have the time difference square in the last term) becomes:

$x(t)=\frac{1}{2}at^2+v_0t+x_0$

And the third equation you posted can be arranged and written as:

$x_f=\dfrac{v_f^2-v_i^2}{2a}+x_i$

These equations are really all you need for problems involving constant acceleration.

i would still need to take in the difference in time for t right.
The way you have re-arranged it makes more sense to me and it looks cleaner too

 

[MarkFL]

i would still need to take in the difference in time for t right.
The way you have re-arranged it makes more sense to me and it looks cleaner too

I simply used $t$ for the elapsed time which is fairly common practice, whereas your professor is having you use $t_2-t_1$. It means the same thing, but your professor's notation is not unusual either.

 

[rashad764]

I simply used $t$ for the elapsed time which is fairly common practice, whereas your professor is having you use $t_2-t_1$. It means the same thing, but your professor's notation is not unusual either.

lets say i was trying to do t2-t1 for x2 and and x1. so 2.5-6=-3.5. how does that work

 

[MarkFL]

lets say i was trying to do t2-t1 for x2 and and x1. so 2.5-6=-3.5. how does that work

I'm not sure what you're asking here, but $t$'s represent time and $x$'s represent distance or position, so you don't want to interchange them.

 

[rashad764]

I'm not sure what you're asking here, but $t$'s represent time and $x$'s represent distance or position, so you don't want to interchange them.

in this problem, what would be the time at position 2 and at position 1

 

[MarkFL]

in this problem, what would be the time at position 2 and at position 1

For the first phase of the problem, we could let:

$t_1=0\text{ s},\,t_2=6\text{ s}\implies t=\Delta t=v_2-v_1=6\text{ s}$

$x_1=0\text{ m},\,x_2=77.4\text{ m}\implies x=\Delta x=x_2-x_1=77.4\text{ m}$

 

[rashad764]

thanks for helping, really appreciate it!