Velocity, Speed, and Time question

In summary, the conversation is discussing a problem where a driver notices that it takes time t to go from one mile marker to the next on a crowded freeway. When the driver increases their speed by 4.5 mph, the time to go one mile decreases by 10 seconds. The problem asks for the original speed of the driver. The conversation goes on to discuss using the equations V=D/T and V2= (D+4.5)/(T-10) to solve for the original speed. It is suggested to use velocity and speed interchangeably and to solve for one of the unknowns in one equation and plug it into the other. Finally, the conversation ends with a summary of the equations and a statement that
  • #1
OUstudent
14
0

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 4.5 mi/h, the time to go one mile decreases by 10 s. What was your original speed?

Homework Equations


V= D/T

The Attempt at a Solution


V1= D/T
V2= (D+4.5)/(T-10)
4.5mph = 6.6ft/s

That's all i could come up with so far. Thanks in advance for the help
 
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  • #2
OUstudent said:

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 4.5 mi/h, the time to go one mile decreases by 10 s. What was your original speed?

Homework Equations


V= D/T

The Attempt at a Solution


V1= D/T
V2= (D+4.5)/(T-10)
4.5mph = 6.6ft/s

That's all i could come up with so far. Thanks in advance for the help

Look at the line I have boldfaced. The question says the speed increases by 4.5 mph, but you're increasing your distance D by 4.5 :eek:
 
  • #3
4.5mph = 6.6ft/s
V1= 1/T

V2+6.6ft/s= (1)/(T-10)

Velocity is used for speed correct?
I also realized it gives you distance.. duhhh..

How do i proceed from here? Am i trying to isolate time to but into the v1 equation?
 
  • #4
OUstudent said:
4.5mph = 6.6ft/s
V1= 1/T

V2+6.6ft/s= (1)/(T-10)

Velocity is used for speed correct?
I also realized it gives you distance.. duhhh..

How do i proceed from here? Am i trying to isolate time to but into the v1 equation?

Yes, here you can use velocity and speed interchangeably.

Notice how the question says "When you increase your speed by 4.5 mi/h" ... it means V2 = V1 + 4.5. How do the equations look now?
 
  • #5
which means T2= t1-10 , I think...

I feel like it should be a simple plug in now but i can't seem to find out how.

V1+6.6= 1/(t1-10)
I'm missing something.. either a way to get time or velocity so i can solve for the other..
 
  • #6
OUstudent said:
which means T2= t1-10 , I think...
Yes, that is correct.
I feel like it should be a simple plug in now but i can't seem to find out how.

V1+6.6= 1/(t1-10)
I'm missing something.. either a way to get time or velocity so i can solve for the other..
I think instead of ft/s, you should stick with miles per hour. Then, you will have two equations in two variables! Do you see it?
 
  • #7
You have two equations and two unknowns. Solve one of the equations for one of the unknowns, and plug the result into the other equation.
 
  • #8
Ok, so what i think you're saying is...

V1+4.5= 1/T1-10

v1= (1/t1-10)-4.5
so.. (1/t1-10-4.5) = 1/t?
then solve that for T?
 
  • #9
OUstudent said:
Ok, so what i think you're saying is...

V1+4.5= 1/T1-10

v1= (1/t1-10)-4.5
so.. (1/t1-10-4.5) = 1/t?
then solve that for T?

The boldface part is not right. What you have in the first line is
[itex]V_{1} = \frac{1}{T_{1}-10} - 4.5[/itex]
and the second line doesn't agree with this. It should read
[itex]\frac{1}{T_{1}} = \frac{1}{T_{1}-10} - 4.5[/itex]
 
  • #10
but don't you plug what i had into the first equation for v1?
i didn't mean to put the parenthesis where i did. i mean to put so. (1/t1-10)-4.5 = 1/t
then solve for t.
is this still not correct?nvm...
 
Last edited:

Related to Velocity, Speed, and Time question

What is the difference between velocity and speed?

Velocity and speed are often used interchangeably, but they have different meanings in physics. Speed is a scalar quantity that measures how fast an object is moving, while velocity is a vector quantity that measures both speed and direction. In other words, velocity tells us not only how fast an object is moving, but also in which direction it is moving.

How is velocity calculated?

Velocity is calculated by dividing the displacement of an object by the time it takes to cover that distance. Mathematically, it is expressed as v = d/t, where v is velocity, d is displacement, and t is time. The SI unit for velocity is meters per second (m/s).

What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time it took to cover that distance. It gives an overall picture of an object's motion. On the other hand, instantaneous velocity is the velocity of an object at a specific moment in time. It is calculated by finding the slope of the object's position-time graph at that point.

How are velocity, speed, and time related?

Velocity, speed, and time are related through the equation v = d/t, where v is velocity, d is displacement, and t is time. This means that if we know any two of these variables, we can calculate the third one. For example, if we know the velocity and time, we can find the displacement; or if we know the displacement and speed, we can find the time.

What is the difference between speed and acceleration?

Speed and acceleration are both related to an object's motion, but they have different meanings. Speed is a measure of how fast an object is moving, while acceleration is a measure of how quickly the object's speed is changing. In other words, acceleration tells us how much an object's speed is increasing or decreasing over time.

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