Trouble with another Kinematics question

In summary, the person rode a bike for 30 minutes, took a water break for 15 minutes, and then rode for another 450 meters at an average velocity of 1.95 meters per second.
  • #1

Homework Statement


A person riding a bike travels at an average velocity of 3m/s for 30min. The person then takes a water break for 15 min. The person finally travels a distance of 450m at an average velocity of 1.5m/s. What is the persons average velocity?

2. Homework Equations
[/B]
  • Average Velocity: Δx/Δt
(Change in positions vs change in time)

  • Conversion

3. The Attempt at a Solution

1.
I first labeled the data.

Trip1

Average Velocity1: 3m/s
for
Time: 30min= 1800s

Break

Time: 15min= 900s

Trip2

Average Velocity2: 1.5m/s
for
Distance: 450m

2.
I drew a picture next of what that would look like
3. Next, I converted the minutes of the first trip and the break taken into seconds
4. I'm assuming that the equation for average velocity must be used in order to solve this.

So here is where I get stuck. I thought of adding the velocities for a total but then thought against it. I then thought to solve for the time in trip two then the distance in trip one.
  • Trip2: T=D/T Time=300s
  • Trip1: D=V⋅T Distance= 5400m
I'm not sure if that's correct. Or if it is, I'm not sure what to do next..
 
Physics news on Phys.org
  • #2
The average velocity is defined as follows, right!

$$v_{av.}=\frac{Total\ Distance}{Total\ time}$$

What will you choose for the total distance and the total time?
 
  • #3
Phylosopher said:
The average velocity is defined as follows, right!

$$v_{av.}=\frac{Total\ Distance}{Total\ time}$$

What will you choose for the total distance and the total time?

I think that the total distance is 5400m added to 450m. I got 5840m as the answer. For times, i added 300s from trip 2 to 1800s from trip1. Then i subtracted 900s from the break and got 1200s total.

I divided those values 5850m/1200s and i got 4.875m/s for the entirety of the trip total. But the question states that the answer is 1.95m/s...
 
  • #4
DracoMalfoy said:
Then i subtracted 900s from the break
Why?
 
  • #5
haruspex said:
Why?

I didnt think that the break was part of the trip i guess. Should that be added to the time?
 
  • #6
DracoMalfoy said:
I think that the total distance is 5400m added to 450m. I got 5840m as the answer. For times, i added 300s from trip 2 to 1800s from trip1. Then i subtracted 900s from the break and got 1200s total.

I divided those values 5850m/1200s and i got 4.875m/s for the entirety of the trip total. But the question states that the answer is 1.95m/s...

You are on the track. But, let us first clear things up.

This might be subjective! But, the question seems to add the break time because it is part of the journey (The total time), this means it should be included, shouldn't?
Moreover, even if you think that the break time is not part of the "journey", then why subtract it?
--Read the question again, carefully.
 
  • #7
Phylosopher said:
You are on the track. But, let us first clear things up.

This might be subjective! But, the question seems to add the break time because it is part of the journey (The total time), this means it should be included, shouldn't?
Moreover, even if you think that the break time is not part of the "journey", then why subtract it?
--Read the question again, carefully.

I guess subtracting it doesn't make sense now that I think about it. I tried adding it in and got 6600s. Then I divided again and got 0.886m/s.
 
  • #8
DracoMalfoy said:
I tried adding it in and got 6600s

Since you now know exactly what to do, try to write everything again from scratch. You will probably not get 6600.
Sometimes it is easier to start from the beginning.
 
  • #9
DracoMalfoy said:
I didnt think that the break was part of the trip i guess. Should that be added to the time?
Yes. But even if not, why subtract it?

By the way, I do take issue with the sloppy wording in the question. Distances and speeds are scalars, displacements and velocities are vectors. To work, the question should refer to speeds, not velocities. Replacing "distance" by "displacement" in the wording instead does not fix it since you would then need to be told that the displacements were in the same direction on the same straight line.
 
  • Like
Likes Phylosopher
  • #10
i
Phylosopher said:
Since you now know exactly what to do, try to write everything again from scratch. You will probably not get 6600.
Sometimes it is easier to start from the beginning.

I think i added the distance to the times by mistake and got 6600 :doh:I got 3000 after adding correctly and did get 1.95m/s. Thanks for the help
 

1. What is Kinematics?

Kinematics is the branch of physics that studies the motion of objects, without considering the causes of the motion.

2. What is the difference between Kinematics and Dynamics?

Kinematics deals with the description of motion, while Dynamics focuses on the causes of motion, such as forces and energy.

3. How do I solve a Kinematics problem?

To solve a Kinematics problem, you need to identify the given information, choose the appropriate kinematic equations, and solve for the unknown variable. It is also important to draw a diagram and label all given and unknown quantities.

4. What are the most common kinematic equations?

The most common kinematic equations are the equations for constant acceleration motion, also known as the SUVAT equations: v = u + at, s = ut + 1/2at^2, v^2 = u^2 + 2as, and s = (u+v)/2 * t.

5. How do I know which kinematic equation to use?

You can determine which kinematic equation to use based on the given information and the unknown variable. For example, if you are given initial and final velocities and acceleration, you can use the equation v = u + at to find the unknown displacement.

Suggested for: Trouble with another Kinematics question

Replies
22
Views
231
Replies
26
Views
517
Replies
30
Views
287
Replies
16
Views
266
Replies
4
Views
766
Replies
3
Views
949
Back
Top