Objects approaching one another with uniform acceleration

[ThatGuyNick]

Homework Statement

persons 1 & 2 are initially at rest (v_i=0m/s) 48m (d) apart. they then run towards each other at constant accelerations
person 1 acceleration = 0.50m/s²
person 2 acceleration = 0.30m/s²

how long until they reach each other/collide (t)?
at the instant they collide, how far has object one gone (xf₁)?

for convenience, we will place the origin at the starting point of the first object
where d=+48m is the initial position of the second object

Homework Equations

x₁=1/2a₁t²
x₂=d+1/2a₂t²

when x₁ = x₂, the players collide at time t = t₁ = t₂

The Attempt at a Solution

1/2a₁t² = d + 1/2a₂t²

note that a₁ = +0.50m/s², while a₂=-0.30m/s² since the first person accelerates in the +x direction and the second person in the -x direction

how to solve for t?

 

[BvU]

Hello Nick, $\qquad$

You bring all terms involving t to one side and all known terms to the other side of the = sign.

It seems to me you are so intimidated by the expression that you don't recognize it as a simple quadratic equation. Unnecessary!

 

[neilparker62]

The persons do not collide when x₁=x₂. I would let x₁=d and x₂ =48-d.

But simplest would be to consider the relative acceleration in which case simply use d=48m for determination of t after which the respective distances should be straightforward.

 

[BvU]

The persons do not collide when x1=x2

Yes they do.

 

[neilparker62]

Yes they do.

True enough - the vector displacements from common reference point must be the same and so apologies to the OP who set up his equations correctly. I was thinking in terms of the scalar distances from separate starting points to the collision point.

 

[Ray Vickson]

Homework Statement

persons 1 & 2 are initially at rest (v_i=0m/s) 48m (d) apart. they then run towards each other at constant accelerations
person 1 acceleration = 0.50m/s²
person 2 acceleration = 0.30m/s²

how long until they reach each other/collide (t)?
at the instant they collide, how far has object one gone (xf₁)?

for convenience, we will place the origin at the starting point of the first object
where d=+48m is the initial position of the second object

Homework Equations

x₁=1/2a₁t²
x₂=d+1/2a₂t²

when x₁ = x₂, the players collide at time t = t₁ = t₂

The Attempt at a Solution

1/2a₁t² = d + 1/2a₂t²

note that a₁ = +0.50m/s², while a₂=-0.30m/s² since the first person accelerates in the +x direction and the second person in the -x direction

how to solve for t?

If you write $t^2 = S$, your equation becomes $0.25 S = 48 - 0.15 S.$ Can you figure out how to get $S?$