How do I finish this?

  • Thread starter thschica
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  • #1
thschica
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A person driving her car at 57 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection .Should she try to stop, or should she make a run for it? The intersection is 15 m wide. Her car's maximum deceleration is -6.6 m/s2, whereas it can accelerate from 57 km/h to 70 km/h in 3.9 s. Ignore the length of her car and her reaction time.

If she hits the brakes, how far will she travel before stopping?
19 m (I know how to do this one )

If she hits the gas instead, how far will she travel before the light turns red?
? m (How would I do this one?)
 

Answers and Replies

  • #2
arildno
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"whereas it can accelerate from 57 km/h to 70 km/h in 3.9 s."
How can you find her possible acceleration from this piece of info?
 
  • #3
Fermat
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her speed changes from 57km/h to 70 km/h, which is a change of 13 km/h, in 3.9s.
13 km/h = 13/3.6 = 3.6 m/s
So, accln is 3.6/3.9 = .925 m/s²
 
  • #4
Fermat
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If she hits the gas instead, how far will she travel before the light turns red?
? m (How would I do this one?)
You know the accln.
You know her initial speed.
You know the time limit during which she can travel.

Use the suvat eqn for distance traveled under constant accln.
 
  • #5
thschica
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a=.93
Initial speed =15.83
time = 2secs
is that right? I just can't seem to get the right answer.
 
  • #6
Fermat
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That's right - you should have used the eqn: s = ut + ½at².

If you used that, what did you get.
 
  • #7
thschica
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Is the correct answer 33.52 ?
 
  • #8
Fermat
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yep - close enough, depending on rounding errors.

a = 0.923 m/s²
u = 15.83 m/s
t = 2 s

s = ut + ½at²
s = 15.83*2 + ½*0.923*4
s = 31.66 + 1.846
s = 33.506 m
========
 
  • #9
thschica
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Thanks a lot for everything!
 

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