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.