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

  • #1

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 + Vot + 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 + Vot
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.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
13,096
3,416
Equation for the person:
X = Xo + Vot
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?
 
  • #3
scottdave
Science Advisor
Homework Helper
Insights Author
1,814
776
You forgot the t square in -6 + 4t = 0.6t^2
 

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

Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
830
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
2
Views
3K
Replies
3
Views
6K
  • Last Post
Replies
2
Views
8K
Top