Determining Time When Comparing Two Equations

  • Thread starter JDK
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  • #1
JDK
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Hello all,

The last time I used these forums for help I got awesome responses. And here I am again needing help. Thought I'd just mosey on over and pop up my question I got. Here it is:

(2) A car and motorcycle start from rest at the same time form rest on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70 m/s2 and the motorcycle accelerates at a uniform rate of 4.40 m/s2.
(a) How much time elapses before the motorcycle overtakes the car?

I know that I need this formula --> d = Vi(t) + 1/2(a)(t^2)
And I know that it should be set up so that the two distance equations (one for the car and one for the motorcycle) are set equal to each other, so I can further solve the junctioned equation for (t). But, I seem to be having some trouble. I need to know how to go about solving this problem. Any help is very much appreciated! Thanks a lot.

- JDK
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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The formula "d = Vi(t) + 1/2(a)(t2)" isn't quite correct. It should be d = Vi(t) + 1/2(a)(t2)+ di where di is the initial distance from some reference point.

In this case, if you take the point from which the motorcycle starts as the reference point, then di= 0 for the motorcycle and di= 25 for the car (since the car starts 25 meters ahead of the motorcycle).
Of course, Vi= 0 for both motorcycle and car.

For the car, a= 3.7 m/s2 so d= (3.7/2)t2+ 25= 1.85 t2+ 25.
For the motorcycle a= 4.4 m/s2 so d= (4.4/2)t2+0= 2.2 t2.

The motorcycle will overtake the car when the two d s are the same:
when 1.85 t2+ 25= 2.2 t2. That's the equation you want to solve.
 
  • #3
JDK
27
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Thank you. Now I understand what I need to do. My problem was not knowing the 'Initial Distance Variable'.

...

Vi(t) + ½(Ac)(t²) + Dic = Vi(t) + ½(Am)(t²) + Dim
(0)(t) + ½(3.70 m/s²)(t²) + 25m = (0)(t) + ½(4.40 m/s²)(t²) + 0m
½ (3.70 m/s²)(t²) + 25m = ½(4.40 m/s²)(t²)
(1.85 m/s²)t² + 25m = (2.2 m/s²)t²

... excuse my lack of knowledge in this area but next am I supposed to square each side of the equation or divide the right side by (1.85)? Or is it something completely different. I seriously must be having an off day... *embarassment* This usually doesn't happen... I've tried a few things on it but I have no idea what the right answer is... that's my problem right now...
 
  • #4
Doc Al
Mentor
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Originally posted by JDK
(1.85 m/s²)t² + 25m = (2.2 m/s²)t²

... excuse my lack of knowledge in this area but next am I supposed to square each side of the equation or divide the right side by (1.85)? Or is it something completely different.
Combine the two terms that contain t2. (Subtract!)
 
  • #5
JDK
27
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*smacks self promptly*

Pheww... I better do a brain check... anyways, so this is what it should look like then - correct? (My symbols are showing up all googled... I'll change that...)

(1.85 m/s²)t² + 25m = (2.2 m/s²)t²
25m = (2.2 m/s²)t² - (1.85 m/s²)t²
25m = (0.35 m/s²)t²
25m / 0.35 m/s² = t²
71.4 s² = t²
(sqroot)71.4 s² = t
8.4 s = t

Thanks by the way! :)
 
Last edited:

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