1-D Kinematics Problem (Running to catch up to a bus)

[Civil_Disobedient]

Homework Statement

A person runs at a constant speed of 4.0 m/s to catch a stationary bus. When she is 6.0m behind it (t=0s), the bus leaves, accelerating with a (constant) acceleration of 1.2 m/s^2. How long does it take her to catch up to the bus?

Homework Equations

X = Xo + volt + 0.5at^2

The Attempt at a Solution

Variables for the person:
Vo = 4
Xo = -6

Variables for the bus:
Vo = 0
Xo = 0
a = 1.2

Equation for the person:
X = Xo + volt
X = -6+4t

Equation for the bus:
X = 0.5at^2
X = 0.5(1.2)t^2
X = 0.6t^2

Setting the two equations equal to each other, I got:
-6 + 4t = 0.6t
-3.4t = -6
t = 5.67s

Not sure what to do from here or if I even did everything correctly up to this point. Thanks for your help.

 

[TSny]

Equation for the person:
X = Xo + volt
X = -6+4t

Equation for the bus:
X = 0.5at^2
X = 0.5(1.2)t^2
X = 0.6t^2

Setting the two equations equal to each other, I got:
-6 + 4t = 0.6t

Should the right hand side be 0.6t²?

Not sure what to do from here or if I even did everything correctly up to this point.

How does the value of time that you get when you solve the equation relate to the time that is asked for in the problem?

 

[scottdave]

You forgot the t square in -6 + 4t = 0.6t^2