# When will these cars pass eachother - kinematics

1. Feb 19, 2006

### teng125

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.
the answer is 3.59m/s^2
thanx

2. Feb 19, 2006

### teng125

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

3. Feb 19, 2006

### phucnv87

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

4. Feb 19, 2006

### teng125

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??

5. Feb 20, 2006

### teng125

pls help.........

6. Feb 21, 2006

### teng125

pls help........

7. Feb 21, 2006

### topsquark

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