Distance traveled correct?(not homework)

In summary, the conversation discusses a self-test in Ch1 of Basic Physics: A Self-Teaching Guide and the difficulties the speaker is having with questions 3 and 5. Question 3 asks for the acceleration of an object traveling at 2 m/s and then at 14 m/s after 3 seconds. The speaker correctly uses the acceleration equation to find the answer of 4 m/s. However, for question 5, the speaker makes a mistake in their calculation of distance traveled during the 3 seconds of accelerated motion. They eventually realize their mistake and correct it, finding the correct answer of 24 m. The summary ends with a note from the speaker to slow down and think carefully to avoid making mistakes in problem solving
  • #1
webslinger67
2
0
I am currently doing the self-test in Ch1 of Basic Physics: A Self-Teaching guide, by Karl F. Kuhn. I am trying to teach myself the basics of physics so I can be ready for March's session of physics class. I had a go at it for 2 days and realized i was going to fail unless I did something to be more comfortable with it.

ok Question 3 says:

3) an object traveling with the speed of 2 m/s. Three seconds later its speed is 14 m/s. How much is its acceleration during this time. Answer= 4 m/s

I get this by using the acceleration equation of a = (V1-V0)/t or (14-2)/3= 4 m/s

now my issue is question 5 which says:

5)How far does the object in question 3 travel during the 3 seconds of its accelerated motion?

so I worked it like this using the acceleration, distance traveled and time of fall relation equation.

d = V0(t)+1/2 * at^2 or 2(1) +1/2(4.67)(3)^2 = 23.02 m

for the V0(t) i used 2m/s times the assumed time of speed after 1 second

for the a (or acceleration) i averaged over 3 seconds the total speed of 14 m/s or 4.67 m/s

the obviously, following the equation, multiplied by the the total time of 3s squared

the book is trying to tell me 24 m is the answer, can someone tell me what I did wrong here, or is the book incorrect?
 
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  • #2
webslinger67 said:
5)How far does the object in question 3 travel during the 3 seconds of its accelerated motion?

so I worked it like this using the acceleration, distance traveled and time of fall relation equation.

d = V0(t)+1/2 * at^2 or 2(1) +1/2(4.67)(3)^2 = 23.02 m

for the V0(t) i used 2m/s times the assumed time of speed after 1 second
In the kinematic equation for distance, t is the time. Why did you put t = 1 in the first term but t = 3 in the second term?? It's given that t = 3.

for the a (or acceleration) i averaged over 3 seconds the total speed of 14 m/s or 4.67 m/s
I have no idea what you're doing here. You already calculated the acceleration to be 4 m/s^2 (not m/s, by the way).

Fix these errors and redo.
 
  • #3
Ok I found my mistake. I averaged the total speed by mistake and used it in place of the acceleration.

here is the correct equation set up.

2(3) +1/2(4)(3)^2=24

I must think too far into the problem, and got to slow down.
 
Last edited:

1. How is distance traveled calculated?

The distance traveled is calculated by multiplying the speed by the time traveled. This equation is represented as distance = speed x time.

2. What is the difference between distance traveled and displacement?

Distance traveled is the total length of the path taken, while displacement is the shortest distance between the starting point and the final position. Displacement takes into account the direction of movement, while distance traveled does not.

3. Does distance traveled depend on the reference frame?

Yes, distance traveled can vary depending on the reference frame used. For example, if an object is traveling in a circular motion, the distance traveled will be different depending on whether the reference frame is fixed or rotating with the object.

4. How does acceleration affect distance traveled?

Acceleration affects distance traveled by changing the speed of the object. The greater the acceleration, the faster the speed and the greater the distance traveled in a given time period.

5. Can distance traveled ever be negative?

No, distance traveled cannot be negative. It is a scalar quantity and represents the magnitude of the displacement, so it is always a positive value.

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