How to find a speed when only a speed and velocity are given?

[DatAshe]

Homework Statement

A car travels along a straight line at a constant speed of 48.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 32.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
mi/hr

Homework Equations

avg=(x_1+x_2+...x_2)/n

The Attempt at a Solution

(48.5+x)/2=32.0
2((48.5+x)/2=32*2
48.5+x=64
(48.5+x)-48.5=64-48.5
x=15.5

This is the only way that I can think of to relate the variables to each other because I don't have a distance or time for any part of in. 15.5 is incorrect, and I'm not looking for an answer, I'm looking for how to relate the variables and find the second constant speed in any situation. Any and all help would be much appreciated. Thank you!

 

[PeterO]

Homework Statement

A car travels along a straight line at a constant speed of 48.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 32.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
mi/hr

Homework Equations

avg=(x_1+x_2+...x_2)/n

The Attempt at a Solution

(48.5+x)/2=32.0
2((48.5+x)/2=32*2
48.5+x=64
(48.5+x)-48.5=64-48.5
x=15.5

This is the only way that I can think of to relate the variables to each other because I don't have a distance or time for any part of in. 15.5 is incorrect, and I'm not looking for an answer, I'm looking for how to relate the variables and find the second constant speed in any situation. Any and all help would be much appreciated. Thank you!

The line highlighted in red above is what you would use if the car traveled at the two separate speeds for equal times. The car actually traveled for the same distance at each speed, so the times are different and the average cannot be calculated in this simple way.

You could always try putting in a numerical value for d; something convenient - perhaps 48.5 miles.
That means the first part of the journey takes 1 hour, and the total distance is 97 miles.
To average 32 mph, the whole journey takes a little over 3 hours - so just over 2 hours to cover the second 48.5 miles, so a speed of approximately 24 mi/h.
[what a pity that first speed wasn't a nice 48 mi/hr.

 

[DatAshe]

Thank you so much! I didn't know that you could do that! You, PeterO, are the bomb!:D

 

[azizlwl]

Average speed= total distance/total time taken.

You know total distance is 2d and total time taken is sum of time of each speed covering a distance d.

Here you have 3 unknowns and you have to find 3 equations to solve the problem.

 

[PeterO]

Thank you so much! I didn't know that you could do that! You, PeterO, are the bomb!:D

you could have an algebraic solution involving d [where the d will eventually cancel out, or substitute a convenient value.
If you are not sure that choosing a particular value will affect/change the situation, you can try two different values, to confirm that the same answer results each time.
Some people find two numerical calculations easier and faster than one algebraic calculation.