Find the original speed of the car

[hgducharme]

Homework Statement

Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 7.7mi/h , the time to go one mile decreases by 15s. What was your original speed?

v₁ = ?
d₁ = 1 mile
t₁ = ?

v₂ = v₁ + 7.7mi/h
d₂ = 1 mile
t₂ = t₁ - 15 seconds

Homework Equations

I'm not sure... We have not covered this in class yet, and we have yet to cover one dimensional kinematics (which is next week). I've seen some equations regarding velocity initial, velocity final, etc., but I don't know how to implement them. The only equations I can come up with are these:

$s_1 = \frac{d_1} {t_1}$
$s_1 = \frac{1 mile} {t_1}$

$s_2 = \frac{d_2} {t_2}$
$s_2 = \frac{1 mile} {t_! - 15 seconds}$

The Attempt at a Solution

This is what I have written on my paper so far:

$\frac{1 mile} {t_1} = \frac{1 mile} {t_1 -15 seconds}$Edit: Umm, not sure where I went wrong with my LaTeX...?

 

[mfb]

The combination of [sub]-tags inside TeX does not work, but I'm not sure if that is the only issue.

$s_1 = \frac{d_1}{t_1}$

Is s1 supposed to be a velocity? That would be an unusual variable name, but okay. You know s2 is not the same as s1, and you even know the differences... include that in your equation, and you can solve it for t1 (for example).

 

[hgducharme]

The combination of [sub]-tags inside TeX does not work, but I'm not sure if that is the only issue.

Is s1 supposed to be a velocity? That would be an unusual variable name, but okay. You know s2 is not the same as s1, and you even know the differences... include that in your equation, and you can solve it for t1 (for example).

Excuse my ignorance, but am I only able to edit once? I can't seem to find the edit button for my original post?

Yes, I apologize for that, s₁ was supposed to be a velocity. I'm going to fully edit my post so the LaTeX will work, make it more clear (e.g. swap out s₁ for v₁, etc.). I also just found the introductory physics formula thread, and I realize that the formula I need is probably in there. I'm going to see what I can do with one of those formulas.

 

[mfb]

Excuse my ignorance, but am I only able to edit once? I can't seem to find the edit button for my original post?

Just for a fixed time (a few hours). I fixed the codes.

You can make a new post with updates formulas.

 

[HalfLostAndSearching]

Sorry about that.

I would first start by defining the variables in the problem and identifying any known information. In this case, we have:

v1 = original speed of the car
v2 = new speed of the car (v2 = v1 + 7.7mi/h)
d1 = distance traveled (1 mile)
t1 = time taken to travel 1 mile at the original speed (unknown)
t2 = time taken to travel 1 mile at the new speed (t2 = t1 - 15 seconds)

Next, I would try to find a relationship between these variables using equations of motion. Since we are dealing with one-dimensional motion, we can use the equation:

v = d/t

Where v is the velocity, d is the distance traveled, and t is the time taken to travel that distance. We can rearrange this equation to solve for t:

t = d/v

Now, we can apply this equation to both scenarios (original speed and new speed):

For the original speed:
t1 = d1/v1

For the new speed:
t2 = d2/v2

Since we know that d1 = d2 = 1 mile, we can set these two equations equal to each other:

t1 = d1/v1 = d2/v2 = t2

Now we can substitute in the values we know:

t1 = 1 mile/v1
t2 = 1 mile/v2

And we also know that t2 = t1 - 15 seconds, so we can set these two equations equal to each other:

1 mile/v2 = 1 mile/v1 - 15 seconds

We can now solve for v1 by isolating it on one side of the equation:

1 mile/v1 = 1 mile/v2 + 15 seconds

v1 = 1 mile/(1 mile/v2 + 15 seconds)

Now, we need to convert the units to be consistent. We know that 1 mile = 1.61 km, so we can convert the velocity units to km/h:

v1 = (1.61 km)/(1.61 km/v2 + 15 seconds/3600 seconds)

v1 = (1.61 km)/(1.61 km/v2 + 0.00417 km/s)

v1 = (1.61 km)/(1.61 km/v2 + 0.00417 km