# Basic Time and Distance Measurement Help

1. Sep 21, 2006

### parwana

Vroom-vroom! As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with a constant acceleration of 8.60 mi/hs. In the adjoining bike lane, a cyclist speeds up from rest to 20.0 mi/h with a constant acceleration of 14.5 mi/hs. Each vehicle maintains a constant velocity after reaching its cruising speed.

(a) For how long is the bicycle ahead of the car?

(b) By what maximum distance does the bicycle lead the car?

for b I tried to find distance of car at Veolocity= 20, distance of bike at veolicty=20 by finding t when velocity =20 and then finding distance by using the result of my equation. My answer, 9.6291 is coming up wrong. Help

2. Sep 21, 2006

Write down the equations of motion. In which relation are the velocities of the cycle and the car in the moment when the cycle isn't ahead of the car anymore?

3. Sep 21, 2006

### parwana

a.

xcar(v=50)= 50^2/2(8.6)= 145.348
tcar(v=50)= 50/8.6= 5.8139

xbike(v=20)= 20^2/2(14.5)= 13.79
tbike(v=20)= 20/14.5= 1.379

145.348 + 50(t-5.8139)= 13.793 + 20(t-1.379)
t= 4.38519

b. x car(v=20)= 20^2/2(8.6)= 23.2558 ft
x bike(v=20)=
t(v=20)= 20/14.5= 1.379
x(t=1.379)= 1/2(14.5)(1.379)^2= 13.786
23.2558-13.786= 9.468

thats wrong as well, what am I doing wrong?

4. Sep 21, 2006

### civil_dude

You have to use correct UNITS. You can't use mph/s for acceleration. You have to convert to feet/s/s and feet/s for velocity!

Last edited: Sep 21, 2006
5. Sep 21, 2006

### parwana

please help more, I have been doing this problem the entire day and havent got it right, I am so fed up

6. Sep 21, 2006

In what units are the results given? Maybe you didn't convert something, as civil_dude stated above. Further on, draw v-t graphs for both the car and the cycle, perhaps the solution will appear more clear.

7. Sep 21, 2006

### parwana

thank u so much guys, especially radou and civil dude, the units were messed up initially, which I fixed, thanks

8. Sep 23, 2006

### destro96

Teacher Couldn't Solve This

My teacher spent 20 minutes in class using a sytem of equations attempting to sovle this w/o success.

Here is how I tried to do it (although I am still not sure if I am on the right track).

Fist, I converted everything to ft/sec

Car: (a)9.0 mi/hr/sec---> 13.1ft/s^2
(vfinal) 50.0mi/hr---> 73.3ft/s

Bike: (a)13.0 mi/hr/sec--->19.3ft/s^2
(vfinal) 20.0mi.hr--->29.3ft/s

I then tried to figure out how big the lead would be once the bike was finished accelerating.

Time for bike to top speed:
3rdKEQ: vf=vi+at
1.52secs second to top speed

Distance for bike to top speed:
4thKEQ: d=vi+vf/2 * t
22.3 ft to top speed.

Now, we need to find where the car is @ 1.538 secs
1stKEQ: d=Vit + 1/2 at^2
7.16 feet

So, at 1.538 secs, the bike is at 22.3 feet and the car is at 7.16 feet.
This puts the bike at its top speed with a 15.13 foot lead on the car.

I just have no idea where to go from here. Am I on the right track?

9. Sep 25, 2006

### civil_dude

Yes, you are on the right track. Now you have the distance between the two and can figure out how long it takes for the bike to travel a distance d, and the car to travel a distance d + 15.13.