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, 201114 yr by Aladdin's

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

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.

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

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

 

Sorry , it's 13.3