Math help

[Natachan]

Perhaps I could use some help with this problem.

When two cars A and B are next to one another they are traveling in the same direction with speeds V(a) and V(b) respectively. If B maintains its constant speed, while A begins to accelerate at a(A), determine the distance d between the two cars the instant A stops.

I got (v^2 - v(a)^2)/2a(A). The book says that's wrong, but I can't figure out how they got their answer.

[Endo]

Perhaps I could use some help with this problem.

When two cars A and B are next to one another they are traveling in the same direction with speeds V(a) and V(b) respectively. If B maintains its constant speed, while A begins to accelerate at a(A), determine the distance d between the two cars the instant A stops.

I got (v^2 - v(a)^2)/2a(A). The book says that's wrong, but I can't figure out how they got their answer.

What's the book's answer? Also, what level math is this? It looks like it could be Physics or some grade of Calculus.

A question, is V(A) implied to be the same as V(B) when they are next to each other? Is there a "T" of any sort in their answer?

[Natachan]

This is dynamics dealing with kinematic a of particles. It's a simple math problem but it is baffling me. It assumes some knowledge of differential equations.

The book has it as the absolute value of (2V(a)V(b)-V(a)^2)/2a(A)

No assumptions about the initial conditions except that they started in the same place. No time variable is in the final answer.

[Endo]

This is dynamics dealing with kinematic a of particles. It's a simple math problem but it is baffling me. It assumes some knowledge of differential equations.

The book has it as the absolute value of (2V(a)V(b)-V(a)^2)/2a(A)

Oh, so that's the shit I get to look forward to in a year.

I will be no help here, having taken neither ODEs or Dynamics.

But, just on a hunch, perhaps because your answer does not take into account the velocity of B? Or, talk to your teacher/tutor about it.

[Surgenator]

I'm going to use laTex syntax for this.

V_{b}(t) = v_b
V_{a}(t) = v_a + a*t

D_{b}(t) = integral (V_{b} dt) = v_b*t
D_{a}(t) = integral (V_{a} dt) = v_a*t + a*t*t/2
\Delta D(t) = D_{b}(t) - D_{a}(t) = (v_b - v_a)*t + a*t*t/2

Assuming 'a' is negative so that car A can come to stop. Car A will come to a stop when V_{a}(t) = 0.
V_{a}(t) = 0 = v_a + a*t_{stop}
after some algebra
t_{stop} = -v_a/a
I'm going to absorbed the negative into 'a'

Plug in the t_{stop} value into the \Delta D equation
\Delta D(t) = (v_b - v_a)*(v_a/a) + a*(v_a/a)*(v_a/a)/2
After a little bit of algebra
\Delta D(t) = (v_b*v_a - v_a*v_a)/a + (v_a*v_a)/(2*a)
\Delta D(t) = 2*(v_b*v_a - v_a*v_a)/(2*a) + (v_a*v_a)/(2*a)
\Delta D(t) = (2*v_b*v_a - 2*v_a*v_a + v_a*v_a)/(2*a)
\Delta D(t) = (2*v_b*v_a - v_a*v_a)/(2*a)
There is your answer

[Losty]

FML no. Just no.

[Natachan]

"Assuming 'a' is negative so that car A can come to stop."

That was my problem, I misread the problem. It was decelerate, not accelerate. Makes it easier to put into the equations of constant acceleration. I was integrating the ads=vdv formula.

[naimless]

Sometimes the books fuck up answers as well.

I never really trusted maths because the poor semantics of the question can often-times fuck up what theory you use.

[Darkstar]

"Assuming 'a' is negative so that car A can come to stop."

That was my problem, I misread the problem. It was decelerate, not accelerate.

Pedantically speaking, deceleration is just acceleration in a negative direction.