Acceleration problem and distance

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?

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?

Andrew Mason
Homework Helper

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?

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

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.

Andrew Mason
Homework Helper
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 $\Delta t$ in terms of $\Delta v$ and a?

AM

I can't solve this...please give me more hints?

Andrew Mason
Homework Helper
I can't solve this...please give me more hints?
Explain how you get 11 seconds.

If $a = \Delta v /\Delta t$ 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

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?
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For Wilma I have 13.7s at constant speed plus 11seconds to include acceleration time is 24.7seconds. Is that right for Wilma?

Andrew Mason