Kinematics

[Aladdin's]

Two cars face each other on a horizontal road.

 

Car A starts from rest at t=0 amd travels with a constant acceleration of 6ft/s^2, until it reaches a speed of 80ft/s. Afterwards it maintain the same speed.Also after t=50 sec , Car B located 6000 ft down the road is traveling towards A with a constant speed of 60 ft/s.

Determine the distance traveled by A when they pass each other.

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I was told that for car B we use (t-50) and for car A (t).

 

My work so far :

 

S(A) = 0.5(6)(t^2) +0 +0

 

S(B) = 6000 + (-60)(t-50) + 0

 

Set S(A)=S(B)

 

I get two values for t , one is accepted t=45.6sec and the other is rejected(negative sign ).

 

Now if I assume that car A meets car B in the first region,

 

V(A) = 0 + (6)(t) => t=13.5 sec

 

Being 45.6 sec > 13.5 , my assumption is wrong ~ Therefore car A meets car B in the second region (Obvious)

 

S(A) = 0.5(6)(13.5)^2 = 530.67 feet

S(B) = 6000 - (60)(13.5 -50) = 8190 feet !! This is impossible , car B is heading towards Car B , so the distance between the two cars must decrease.

 

We've answered a similar question in class , but the difference was that the two cars was launched at the same time ....

 

Thanks for standing by ,

Regards,

 

 

Edit: Error in units.

Edited April 2, 2011 by Aladdin's

[swansont]

Are the units as you have given them, i.e. really mixed between SI and English?

[Aladdin's]

Are the units as you have given them, i.e. really mixed between SI and English?

 

 

Oh sorry for this error , all units are in the English system. I'll edit my post, thanks.

[swansont]

If a = 60ft/sec^2, the car reaches 80 ft/sec in 1.33 seconds. If it's 6 ft/sec^2, the time is 13.3 seconds. I don't see where you're getting 13.5

[Aladdin's]

I've calculated that without the use of a calculator.

 

Sorry , it's 13.3