# Acceleration and distance problem?

1. Sep 30, 2009

### 1irishman

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
I'm not really sure, but i'm guessing these ones:

v=d/t
d=vit + 1/2 at^2
is this a vector problem maybe?

3. The attempt at a solution
i'm not sure how to approach this at all...can i have a couple of hints please? Thank you :-)

2. Sep 30, 2009

### willem2

you'll need to find d(t) for Fred and wilma both while they accelerate and when they run
at constant speed, and then you have to find out when d_fred(t) = 100 and when d_wilma(t)=100

3. Sep 30, 2009

### 1irishman

i dont know how to do that...

are these values for fred and wilma right?

fred

vi=0
a=0.500m/s^2
vf=5s
d=100

wilma

vi=0
a=
vf=6.00s
d=100

4. Oct 1, 2009

### willem2

you can find a for wilma with: v_f = v_i + at
these parameters are for the first phase of fred and wilma's movement when they
accelerate and will not reach the bus stop yet anyway. you want to find out where they are at the end of this phase and at what time. d=100 doesn't make any sense yet

After accelerating they will reach the bus stop running at constant speed, so you can
use speed is distance/time to find out how long the second phase of running took

5. Oct 1, 2009

### 1irishman

okay, i have calculated 'a' for wilma...not sure if it's right though:
using this formula to get 'a' >>>> vf=vi+at
6.00=a(6.00)
a=1m/s^2
now, i have calculated 'd' for wilma...again not sure if it's right though:
using this formula to get 'd' >>>> d=vit+1/2at^2
d=1/2(1)(6)^2
=1/2(36)
d=18m
Am i on the right track here?

-------------------

6. Oct 1, 2009

### 1irishman

I have calculated 'd' for Fred...i think it's right but not sure though:
using this formula to get 'd' >>>>> d=vit+1/2at^2
d=1/2(0.500)(5)^2
=1/2(12.5)
d=6.25m
Am i on right track here?

7. Oct 1, 2009

### 1irishman

since they both run 100m, i said that Fred had 93.75m left so because constant speed i used v=d/t and t=d/v so 93.75/5 = 18.75 + 11s so far = 29.75...i checked the book and this answer was wrong...i don't understand why?
I used the same method as above for Wilmas distance and the answer was right at 24.7s.

8. Oct 1, 2009

### willem2

you have the time that Fred acccelerates wrong. you need to use vf=vi+at
to find out for how long Fred accelerates. Fred also doesn't wait for 5 seconds before
starting

9. Oct 1, 2009

### 1irishman

Okay, now 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?

10. Oct 1, 2009

### 1irishman

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.

11. Oct 1, 2009

### willem2

where do you get those 11 seconds from?

12. Oct 2, 2009

### 1irishman

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

13. Oct 2, 2009

### willem2

why do you think this takes 11 seconds?. Fred doesn't have to stop accelerating and start
moving at constant speed at the same time as Wilma

14. Oct 2, 2009

### 1irishman

so one time is 5s and the other is 6s right?

15. Oct 2, 2009

### 1irishman

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?

16. Oct 2, 2009

### RoyalCat

If Fred's acceleration is $$0.5 \tfrac{m}{s^2}$$, it takes him longer than 5 seconds to accelerate to his final velocity of $$5 \tfrac{m}{s}$$

Split Fred and Wilma's motions into two regimes.
One where they are accelerating, and one where they are traveling with a constant velocity.

Use the kinematic formulas for motion under constant acceleration to find the time it takes them to travel under each of the regimes, and the distances they cover:

$$v(t)=v_0+at$$

$$v^2=v_0^2+2ad$$

$$d(t)=d_0+v_0 t+\tfrac{1}{2}at^2$$

For the second part of their motion, they're traveling under constant velocity, for which the following holds true:

$$a=0$$

$$d(t)=d_0+v_0 t$$

17. Oct 2, 2009

### willem2

Yes those are both right

18. Oct 2, 2009

### 1irishman

okay thank you. so the 11 seconds was for wilma only right?

19. Oct 2, 2009

### willem2

the 11 seconds for wilma came for the 5 seconds wait added to her 6 seconds of acceleration.

20. Oct 2, 2009

### 1irishman

okay super, thanks for your help willem2!