# When will these cars pass eachother - kinematics

Car A with u=0/s is on the northbound land of a highway and car B is travelling in th southbound lane at constant u=26.67m/s.At t=0 , A starts and accelerates at a constant rate a(A) . while at t=5 , B decelerates with constant magnitude of a(A)/6 . knowing that when the two carspass each other at x=90m , and v(A) = v(B),determine acceleration of car A.

i have tried it many times but stil can't ind the answer.pls help.
thanx

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oh...the length between the two cars is d in which we have to find in the second part

You just need to write the equations of velocity and movement of each car. Then you'll find out the answer

for v(A)=180a(A)--------1
s(A)=1/2 at^2--------2

v(B)=26.67 - 0.5a(B)t--------3
d-90=26.67t - 0.5a(B)t^2-----4

is it rite??

pls help.........

pls help........

teng125 said:
pls help........
Teng! I'm not a moderator but all you have to do is ask ONCE! :surprised

Tricky! I like it.

In the following, I have defined + to be Northward.
For car A: Car A starts at position $$x_{0A}$$ at a speed of $$v_{0A} = 0m/s$$ with an acceleration of a(A).
$$x_A=x_{0A}+v_{0A}t+(1/2)a_At^2$$; $$v_A=v_{0A}+a_At$$
$$x_A=x_{0A}+(1/2)a(A)t^2$$; $$v_A=a(A)t$$

For car B: Car B starts at position $$x_{0B}$$ at a speed of $$v_{0B} = -26.67 m/s$$ with an acceleration of -(1/6)a(A).
$$x_B=x_{0B}+v_{0B}t+(1/2)a_Bt^2$$; $$v_B=v_{0B}+a_Bt$$
$$x_B=x_{0B}-26.67t-(1/12)a(A)t^2$$; $$v_A=-26.67-(1/6)a(A)t$$

Now, at x=90 m, both cars pass each other and have the same speed. We don't have a specific equation to find t or a(A), but we have two equations we can write. First, at time t both cars have the same speed. Thus at (90 m, t):
$$a(A)t=-26.67-(1/6)a(A)t$$

Second, both cars are at the same point at this time, so:
$$x_{0A}+(1.2)a(A)t^2=x_{0B}-26.67t-(1/12)a(A)t^2 =90 m$$
(This last is actually THREE equations. Both distances are equal to 90 m as well as each other.)

You've got four equations in four unknowns, so you should be able to solve the system.

-Dan