Physics problems

Just two physics problems involving kinematics. I have the answers but I don't know how to get to them. Any help would be appreciated.

The problems:

1) An object moving at constant acceleration covers two equal, consecutive, 11 m long segments, the first one in t1 = 0.93s and the second one in t2 = 2.05s . What was the speed of the object (in m/s) at the beginning of the first segment?

Answer: 15.5

2) A terain vehicle, from its intial position, point A should reach a point D which is d= 12.8 km away from the straight road on which the vehicle is on. The point on the road closest to the point D is the point C. The speed of the vehicle while on the road is Vr= 56 km/h while the speed of the vehicle outside of the road is Vt= 34 km/h. Assuming that the vehicle will move along a straight line from point B where it leaves the road, calculate the distance BC so that the overall time ABD is minimum.

Answer: 9.8

I know the the wording is kinda shaky but I just copy pasta'ed it.

For 1, the equation you need is d=vt+.5at^2. Just fill it in:
11=v*t1+.5*a*t1^2
11+11=v*(t1+t2)+.5*a*(t1+t2)^2
Rewrite one of them in terms of .5a=something, then replace .5a in the other with that something and solve the resulting equation for v.

For 2, make an equation that expresses the time for ABD in terms of BC (and AC, I think, since it's not given). Then get the derivative, set to zero, solve for BC.

speed-time graph etc..

Vectors?

2nd question is impossible. You need to know either AC or AB to solve it.

OP here, first question the answer that the prof gave is actually wrong. I used what #2 said (I slapped myself later for not seeing that and thinkng too hard) and got another physics prof to verify my answer.

Second question.. I have no idea. It looks like it can be solved like any related rates question except like #5 said... I need one more side. Is there another way to do it?

>>5
The minimum is independent of AC.

For second question:

distance from B to C: t

time of path: (t-A)/56  +  (sqrt((12.8)^2 + t^2))/34

derivative: 1/56 + 1/34 * ((12.8)^2 + t^2)^(-1/2) * 1/2 * 2t
 = 1/56 + 1/34 * ((12.8)^2 + t^2)^(-1/2) * t

Set it equal to 0:

-34/56 = (163.84 + t^2)^(-1/2) * t

1156/3136 = t^2/(163.84 + t^2)

t^2 - t^2 * (1156/3136) = 163.84 * (1156/3136)
t^2 * (1980/3136) = 163.84 * (1156/3136)
t^2 * 1980 = 163.84 * 1156
t = 9.78..

>>7,8
oh yeah...i was looking at AB/50 + BD/34 = T and trying to figure out how AB is dependant on x...

>>9
err...dependant on CD and x, without AC coming in anywhere.

>>8
(t-A)
Wouldn't that be (A-t)? (Assuming by A you mean AC.)
You can see this when you try to fill in your result here:
-34/56 = (163.84 + t^2)^(-1/2) * t
The left side is negative, but the right side is positive with a positive t, so they can't be equal. In fact, when you square that equation you should add the condition that both sides are of equal sign. The result you'd then get is t=-9.78..., an important difference. If you use (A-t) at the start, you do get t=9.78... which is the correct solution.

should add the condition that both sides are of equal sign.
Actually, no, I'm wrong. The real mistake is at the end where you neglect that t^2=... has two solutions, and going back to the earlier equations you can see that the negative one is correct. (Whereas with (A-t) it would be the positive one.)

>>12
Yeah, (A-t) is correct - sorry about that.

well if you want to be totally illogical it could be abs(A-t)

speed-time graph faggots

it's the simplest piece of piss inthe world

>>15
olol, making a graph for a simple calculation.
Are you an American, by any chance?

Chrono Trigger Fans Unite, Bitches!

>>16
stfu n00b, you visualise the graph to quickly determine the equation needed

OR JUST DRAW THE FRICKING THIS LIKE THIS

A___________________________B______C
                             \     |
                              \    |
                               \   |
                                \  |
                                 \ |
                                  \|
                                   D

>>19
DRAW THAT FUCKING THIS

THISSY