Basic Time and Distance Measurement Help

[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

 

[radou]

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?

 

[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

Thats my answer but its not right, please help

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?

 

[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!

 

[parwana]

please help more, I have been doing this problem the entire day and haven't got it right, I am so fed up

 

[radou]

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.

 

[parwana]

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

 

[destro96]

Teacher Couldn't Solve This

My teacher spent 20 minutes in class using a system 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?

 

[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.