Distance traveled correct?(not homework)

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SUMMARY

The discussion focuses on calculating the distance traveled by an object under acceleration, specifically using the kinematic equations. The user initially misapplied the acceleration and time variables, leading to an incorrect distance calculation of 23.02 m instead of the correct answer of 24 m. The correct formula is d = V0(t) + 1/2 * at², where V0 is the initial velocity, a is the acceleration, and t is the total time. The user ultimately identified their mistake in averaging the speed instead of using the calculated acceleration of 4 m/s².

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webslinger67
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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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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.
 
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:

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