Approaching vehicles with acceleration

[JohnnyGui]

Homework Statement

Two vehicles, A and B, are approaching each other. Vehicle A has a constant velocity of $v_A$. Vehicle B has a constant acceleration $a$ with an initial velocity of $v_B$. The starting distance at $t=0$ between them is $D$.

Is there a way to solve at what time $t$ they'd meet?

Homework Equations

$D = v_At + v_Bt + \frac{1}{2} \cdot a \cdot t^2$

The Attempt at a Solution

I deduced the equation above but I can't find a way to solve for t. Simplifying gives:

$\frac{D}{t} = v_A + v_B + \frac{1}{2} \cdot a \cdot t$

Is there a way to solve for t without plotting these two equations?

 

[Ray Vickson]

Homework Statement

Two vehicles, A and B, are approaching each other. Vehicle A has a constant velocity of $v_A$. Vehicle B has a constant acceleration $a$ with an initial velocity of $v_B$. The starting distance at $t=0$ between them is $D$.

Is there a way to solve at what time $t$ they'd meet?

Homework Equations

$D = v_At + v_Bt + \frac{1}{2} \cdot a \cdot t^2$

The Attempt at a Solution

I deduced the equation above but I can't find a way to solve for t. Simplifying gives:

$\frac{D}{t} = v_A + v_B + \frac{1}{2} \cdot a \cdot t$

Is there a way to solve for t without plotting these two equations?

It is just a quadratic equation in $t$, so is easily solved using the standard quadratic-solution formulas.

 

[PeroK]

Your equation looks very like a quadratic in $t$ to me!

 

[JohnnyGui]

Your equation looks very like a quadratic in ttt to me!

It is just a quadratic equation in ttt, so is easily solved using the standard quadratic-solution formulas.

Ah, for some reason I thought this wasn't possible because there are two parameters with either a $t$ or a $\frac{1}{t}$ in it. But after looking at it I noticed it could be written as:

$\frac{1}{2} \cdot at^2 + (v_A + v_B)t - D = 0$

And then use the abc-formula in which $a=\frac{1}{2}\cdot a$ and $b=(v_A + v_B)$ and $c = D$. Is this correct?

 

[PeroK]

Ah, for some reason I thought this wasn't possible because there are two parameters with either a $t$ or a $\frac{1}{t}$ in it. But after looking at it I noticed it could be written as:

$\frac{1}{2} \cdot at^2 + (v_A + v_B)t - D = 0$

And then use the abc-formula in which $a=\frac{1}{2}\cdot a$ and $b=(v_A + v_B)$ and $c = D$. Is this correct?

You already had the quadratic in section 2 of your original post.

I guess you mean $c = -D$?

You may like to think what difference it makes if vehicle A is at rest and vehicle B is initially moving at the combined speed $v_a + v_b$.

 

[JohnnyGui]

You already had the quadratic in section 2 of your original post.

I guess you mean $c = -D$?

You may like to think what difference it makes if vehicle A is at rest and vehicle B is initially moving at the combined speed $v_a + v_b$.

Apologies, I did mean $c = -D$. I take it you mean that I could just put the equation in section 2 in parentheses? My initial target was to use the abc-formula to only give the positive solution of $t$.