Acceleration problem and distance

  • Thread starter 1irishman
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  • #1
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Homework Statement


Fred and Wilma leave their house at the same time heading for the bus stop 100m away. Fred starts at rest, accelerates at 0.500 m/s^2 up to a maximum speed of 5.00m/s, which he holds as he runs for the bus stop. Wilma gives Fred a 5.00 second head start and then accelerates for 6.00 seconds up to her maximum speed of 6.00m/s. Who gets to the bus stop first, and how long do they wait for the second person to arrive?


Homework Equations


v=d/t
d=vit + 1/2 at^2
vf=vi+at


The Attempt at a Solution


i have Fred accelerates for 10 seconds. and in 10 seconds he covered a distance of 25m. plus his acceleration distance of 6.25m = 31.25m total. 100m minus 31.25m is 68.75m remaining distance at constant speed so t=d/v so 68.75/5 = 13.75 s plus 11 seconds is 24.75. The answer in the book for Fred's time says 25seconds and the answer in the book for Wilma says 24.7seconds. Am i right here? It feels like i'm getting closer to the answer unless i've already got it?
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement


Fred and Wilma leave their house at the same time heading for the bus stop 100m away. Fred starts at rest, accelerates at 0.500 m/s^2 up to a maximum speed of 5.00m/s, which he holds as he runs for the bus stop. Wilma gives Fred a 5.00 second head start and then accelerates for 6.00 seconds up to her maximum speed of 6.00m/s. Who gets to the bus stop first, and how long do they wait for the second person to arrive?


Homework Equations


v=d/t
d=vit + 1/2 at^2
vf=vi+at


The Attempt at a Solution


i have Fred accelerates for 10 seconds. and in 10 seconds he covered a distance of 25m. plus his acceleration distance of 6.25m = 31.25m total.
Why are you adding 6.25 m? You have already determined his acceleration distance is 25m.

100m minus 31.25m is 68.75m remaining distance at constant speed so t=d/v so 68.75/5 = 13.75 s plus 11 seconds is 24.75. The answer in the book for Fred's time says 25seconds and the answer in the book for Wilma says 24.7seconds. Am i right here? It feels like i'm getting closer to the answer unless i've already got it?
The answer in the book is right.

AM
 
  • #3
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Fred's acceleration distance is 25metres. So i got 100m-25m=75m left for Fred plugged into t=d/v=75/5=15 seconds plus 11seconds equals 26seconds. Why am i one second off the correct answer of 25seconds? Please help? thanks.
 
  • #4
Andrew Mason
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Fred's acceleration distance is 25metres. So i got 100m-25m=75m left for Fred plugged into t=d/v=75/5=15 seconds plus 11seconds equals 26seconds. Why am i one second off the correct answer of 25seconds? Please help? thanks.
How do you get 11 seconds? What is [itex]\Delta t[/itex] in terms of [itex]\Delta v[/itex] and a?

AM
 
  • #5
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I can't solve this...please give me more hints?
 
  • #6
Andrew Mason
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I can't solve this...please give me more hints?
Explain how you get 11 seconds.

If [itex]a = \Delta v /\Delta t[/itex] and a = .5 m/s^2, vf = 5 m/s and vi = 0 you should be able to work out the time. It is not 11 seconds.

AM
 
  • #7
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okay...i have t=10s for Fred to accelerate, and 15s at constant speed for Fred total time is 25seconds. Is that right for Fred?
-------------------
For Wilma I have 13.7s at constant speed plus 11seconds to include acceleration time is 24.7seconds. Is that right for Wilma?
 
  • #8
Andrew Mason
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okay...i have t=10s for Fred to accelerate, and 15s at constant speed for Fred total time is 25seconds. Is that right for Fred?
-------------------
For Wilma I have 13.7s at constant speed plus 11seconds to include acceleration time is 24.7seconds. Is that right for Wilma?
Your book says that is the right answer.

AM
 

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