How do I isolate dT from: 1/2(aB)(dT)^2=1/2(aA)(VA)+dT(VA)? Details have specific info. Spent hours, need help

Question

How do I isolate dT from: 1/2(aB)(dT)^2=1/2(aA)(VA)+dT(VA) See details for more info. Desperately need help!!

"Car A and B start side by side. At t=0 A begins to accelerate at a constant rate, which continues for time tA unti it reaches speed vA, at which it continues travelling at constant speed vA. At the same moment A reaches speed vA, B takes off after A, accelerating at a constant rate aB. At t = t*, car B passed car A."

In an attempt to solve the above question to find (t*- tA), or (dT), I have the following formula, AVOIDING to use t* or tA because that would make the equation more complex:

position of B at any point of time: 1/2(aB)(dT)^2
position of A at any point of time (since the passing of car B is after tA): 1/2(aA)(VA)+ dT(VA)

Then the point of crossing would be:
1/2(aB)(dT)^2=1/2(aA)(VA)+dT(VA)

However, I can't isolate dT to determine the time elapsed from the moment B begins to move. I need to know how to proceed or where's a mistake. I'm spent and exhausted.

Answer 1

I agree with grsym's analysis, but think that his last line has a minor error. The last 3 lines of his answer should read as follows:

aB(dT)^2 - 2vA(dT) - aA(tA)^2 = 0

the solution of which is

dT = {vA + sqrt[(vA)^2 + aAaB(tA)^2]}/aB

(Note that the quadratic formula gives TWO solutions to a quadratic equation. The OTHER solution would involve SUBTRACTING the square root term. We can see that this can't possibly give the right value for dT, because it would result in a time LESS than the amount of time it takes B to reach A's velocity ... and B can't possibly catch up with A while still going slower than A. Therefore, the root shown above is the meaningful result. The other root is extraneous.)

Answer 2

I'll assume that you mean delta t and not dt (second order differentials and such don't work like that). If this is so, take everything and put it on one side, group all the terms with dt^2, dt, and no dt and use the quadratic formula to solve for dt.

If it really is differentials you're playing with, you might want to look for a different method to solve the problem.

Answer 3

Distance travelled by A after time t* is sA given by
sA = (0.5)aA(tA)^2 + (vA)dT

Distance travelled by B after time t* is sB given by
sB = (0.5)aB(dT)^2

These two distances are equal so equating the two right-hand sides of the equations and multiplting through by 2 we get the quadratic equation

aB(dT)^2 - 2vA(dT) - aA(tA)^2 = 0

the solution of which is

dT = {vA + sqrt[(vA)^2 + aAaB(tA)^2]}/2

Answer 4

No need for the rest of the explanation. It's simple algebra.

Just take everything that's being done to it on the left and do the oppisite to the expression on the right.

Answer 5

Check to make sure you wrote the problem correctly. I am having problems with it too.