Easy velocity/acceleration problem, could use some help please

  • Thread starter buyerat
  • Start date
In summary, a backpacker with an average velocity of 1.21 m/s, due west, reaches her destination after hiking for 6.44 km with an average velocity of 2.54 m/s, due west, and then turning around and hiking with an average velocity of 0.470 m/s, due east. To find the distance she walked east, the equation is set up as (6440 - d)/(6440/2.6 + 2.35*d) = 1.33, where d represents the distance walked east. After solving for d, the distance walked east is determined to be approximately 2,143.35 meters.
  • #1
buyerat
11
0

Homework Statement



In reaching her destination, a backpacker walks with an average velocity of 1.21 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.54 m/s, due west, turns around, and hikes with an average velocity of 0.470 m/s, due east. How far east did she walk?

Homework Equations



V = Xf - Xo
T


The Attempt at a Solution


From what I understand the time it takes to go west 6440 plus the time it takes to go east = the total time it takes to go the full distance. I believe you set it up like

6440-X2 6440 X2
_____ = _____ + _____

0.470 2.54 0.470


I'm not sure if this is wrong or my arithmetic is just failing me. Any help would be great.

Thanks
 
Physics news on Phys.org
  • #2
See the triple777's post.
 
  • #3
I was just looking it over but I do not understand where you pulled 2.35 from
 
  • #4
buyerat said:
I was just looking it over but I do not understand where you pulled 2.35 from
It is 1/0.425
 
  • #5
Ok I don't really see why you do that, but I can figure it out hopefully. At the end you write the equation once as The average velocity 1.33 = (6440 - d)/(6440/2.6 + 2.35*d)
then at the end you said 1.33 = (6440 - d)/(2477 - 2.35d)

Is there some reason the signs are switched before the 2.35d? I'm not really sure how to do the algebra at the end either with moving it around if any help could come there.
I have something like 1.21 = (6440 - d)/(2535.43 + or minus i don't know 2.13d).
 
  • #6
buyerat said:
Ok I don't really see why you do that, but I can figure it out hopefully. At the end you write the equation once as The average velocity 1.33 = (6440 - d)/(6440/2.6 + 2.35*d)
then at the end you said 1.33 = (6440 - d)/(2477 - 2.35d)

Is there some reason the signs are switched before the 2.35d? I'm not really sure how to do the algebra at the end either with moving it around if any help could come there.
I have something like 1.21 = (6440 - d)/(2535.43 + or minus i don't know 2.13d).
It should be +2.35*d. If it is -ve it is typo.
 
  • #7
I really can't seem to get the right answer if you wouldn't mind helping further. I seem to be coming up with 2143.35m, but that does not seem to be the answer.
 
  • #8
I got it now thank you very much for your help.
 

FAQ: Easy velocity/acceleration problem, could use some help please

1. What is the difference between velocity and acceleration?

Velocity is a measure of an object's speed and direction of motion, while acceleration is the rate of change of an object's velocity over time. In other words, velocity tells us how fast an object is moving, and acceleration tells us how much its velocity is changing.

2. How do I solve an easy velocity/acceleration problem?

To solve an easy velocity/acceleration problem, you first need to identify the given information, such as initial and final velocities, time, and acceleration. Then, you can use the appropriate equations, such as v = u + at for constant acceleration, to find the unknown variable.

3. What are some common units used for velocity and acceleration?

The most common unit for velocity is meters per second (m/s) or kilometers per hour (km/h), while the most common unit for acceleration is meters per second squared (m/s^2) or kilometers per hour squared (km/h^2).

4. Can you give an example of an easy velocity/acceleration problem?

Sure, an example of an easy velocity/acceleration problem would be: A car is traveling at a constant speed of 50 km/h. After 2 hours, the car's speed increases to 70 km/h for the next 3 hours. What is the car's average acceleration? In this case, we can use the equation a = (v - u)/t, where v = 70 km/h, u = 50 km/h, and t = 3 hours. Plugging in the values, we get a = (70 km/h - 50 km/h)/3 h = 20 km/h / 3 h = 6.67 km/h^2.

5. Are there any shortcuts or tricks for solving velocity/acceleration problems?

While understanding the concepts and equations for velocity and acceleration is important, there are some shortcuts or tricks that can help with solving problems quickly. For example, if acceleration is constant, the average velocity can be found by taking the average of the initial and final velocities. Also, if the time is not given, you can use the equation v^2 = u^2 + 2as to find the final velocity without needing to know the time.

Similar threads

Basic Kinematics Problem -- Backpacker's average velocity and distance....
Replies
3
Views
2K
New Velocity/Distance Question 1/25
Replies
4
Views
1K
In reaching her destination, a backpacker walks with an average velocity of 1.21
Replies
4
Views
4K
Momentum: instantaneous or average?
Replies
9
Views
320
Runner's Average Velocity: 0.014 m/s
Replies
17
Views
1K
A difficult 1D kinematics problem dealing with displacement and velocity
Replies
3
Views
7K
What is the distance the backpacker walked east?
Replies
4
Views
2K
How Far East Did the Backpacker Walk?
Replies
3
Views
2K
Hiker Displacement: Find Distance East
Replies
2
Views
2K
Vectors, how to find average acceleration
Replies
13
Views
6K

Hot Threads

Recent Insights

Back
Top