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I'm not sure how to go about this problem. I don't want the numerical answer, just help with how to approach it.

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- #1

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I'm not sure how to go about this problem. I don't want the numerical answer, just help with how to approach it.

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cepheid

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You haven't really stated what the problem is. What are trying to determine?

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cepheid

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The motorcycle has an initial position and initial velocity of 0. Its motion is divided into two different phases, one with constant acceleration and then another with constant velocity (you have to figure out when this second phase starts up).

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cepheid

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Well, we have no way of knowing what you are doing wrong unless if you show us your work. Post your solution here.

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[tex]v_{0}= 0 km/s[/tex]

[tex]a=0.0042 km/s^{2}[/tex]

[tex]v= 108/3600 km/s[/tex]

[tex]delta x= 1.1 km[/tex]

[tex]108/3600 = 0 + 0.0042(t)[/tex]

[tex]t=108/15.12 for the time of acceleration[/tex]

[tex]delta x=(1/2)(0.0042(108/15.12)^{2}[/tex]

[tex]delta x=0.1071428571 km[/tex]

[tex]1.1 km - 0.1071428571 km = 0.9928571429 km[/tex]

[tex]0.9928571429 km = (108/3600)*t[/tex]

[tex]t=19.85714286 s[/tex]

[tex]19.85714286+(108/15.12)=27s[/tex]

[tex]a=0.0042 km/s^{2}[/tex]

[tex]v= 108/3600 km/s[/tex]

[tex]delta x= 1.1 km[/tex]

[tex]108/3600 = 0 + 0.0042(t)[/tex]

[tex]t=108/15.12 for the time of acceleration[/tex]

[tex]delta x=(1/2)(0.0042(108/15.12)^{2}[/tex]

[tex]delta x=0.1071428571 km[/tex]

[tex]1.1 km - 0.1071428571 km = 0.9928571429 km[/tex]

[tex]0.9928571429 km = (108/3600)*t[/tex]

[tex]t=19.85714286 s[/tex]

[tex]19.85714286+(108/15.12)=27s[/tex]

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- #8

verty

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You should work in meters and seconds, it is less confusing.

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cepheid

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Sorry for not responding sooner. I've been busy.

Good. However, as it has been pointed out, it's probably better if you just consistently use one system of units. In this case, I would convert everything into standard SI units, a.k.a. mks (metre-kilogram-second) units.

By the way, I should mention that you really didn't need the full power of LaTeX for most of these equations. Try using the [ SUB ] [/ SUB ] and [ SUP ] [/ SUP ] tags in order to make regular text subscripted and superscripted respectively (without spaces between the square brackets and the tag text). If you use the buttons marked X^{2} and X_{2} above the reply box, it will insert these tags around highlighted text in the reply box automatically. That having been said, I did fix up some of the LaTeX in the quotes of your post. You can click on my versions to see how the source code differs from yours.

Yes, this is the time required for the motorcycle to accelerate up to 108 km/h (30 m/s). However, it is not the total travel time. The rest of the time is spent travelling at a constant speed. You should probably call this time interval t_{1} or something, to identify it.

This part is also good. You've figured out how far the motorcycle travels while accelerating. The rest of the distance has to be covered at a constant speed.

This calculation step is right, but the answer you got is totally wrong. Try this calculation again. Let's call the result t_{2}

Yes, t_{1} + t_{2} is the total travel time of the motorcycle when it is catching up. Hopefully, once you've calculated t_{2} correctly, you'll get the right answer.

When you work out the speed of the car, don't forget that the car has been travelling 1.95 s longer than the motorcycle.

[tex]v_{0}= 0~km/s[/tex]

[tex]a=0.0042~km/s^{2}[/tex]

[tex]v= 108/3600~km/s[/tex]

[tex]\Delta x= 1.1~km[/tex]

Good. However, as it has been pointed out, it's probably better if you just consistently use one system of units. In this case, I would convert everything into standard SI units, a.k.a. mks (metre-kilogram-second) units.

By the way, I should mention that you really didn't need the full power of LaTeX for most of these equations. Try using the [ SUB ] [/ SUB ] and [ SUP ] [/ SUP ] tags in order to make regular text subscripted and superscripted respectively (without spaces between the square brackets and the tag text). If you use the buttons marked X

[tex]108/3600 = 0 + 0.0042(t)[/tex]

[tex]t=108/15.12~\textrm{for the time of acceleration}[/tex]

Yes, this is the time required for the motorcycle to accelerate up to 108 km/h (30 m/s). However, it is not the total travel time. The rest of the time is spent travelling at a constant speed. You should probably call this time interval t

[tex]\Delta x=(1/2)(0.0042(108/15.12)^{2}[/tex]

[tex]\Delta x=0.1071428571~km[/tex]

[tex]1.1~km - 0.1071428571~km = 0.9928571429~km[/tex]

This part is also good. You've figured out how far the motorcycle travels while accelerating. The rest of the distance has to be covered at a constant speed.

[tex]0.9928571429~km = (108/3600)t[/tex]

[tex]t=19.85714286~s[/tex]

This calculation step is right, but the answer you got is totally wrong. Try this calculation again. Let's call the result t

[tex]19.85714286+(108/15.12)=27~s[/tex]

Yes, t

When you work out the speed of the car, don't forget that the car has been travelling 1.95 s longer than the motorcycle.

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