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

• Civil_Disobedient
In summary, this conversation discusses a problem where a person is trying to catch a bus that is initially stationary and then accelerates at a constant rate. By setting the equations for the person and the bus equal to each other, with the person's distance being 6.0m behind the bus at t=0s, the solution is found to be t = 5.67s.
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 + 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.

Civil_Disobedient said:
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?

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

## What is 1-D kinematics?

1-D kinematics is a branch of physics that deals with the motion of objects along a straight line, without considering any forces acting on the object.

## What is a "catch-up" problem in 1-D kinematics?

A "catch-up" problem in 1-D kinematics is a type of problem where an object starts from a certain position and moves along a straight line to catch up to another moving object that started moving earlier.

## How do you solve a "catch-up" problem in 1-D kinematics?

To solve a "catch-up" problem in 1-D kinematics, you can use the formula: distance = rate x time. You will need to set up two equations, one for each object, and solve for the time when the distances are equal.

## What factors can affect the outcome of a "catch-up" problem?

The outcome of a "catch-up" problem can be affected by factors such as the initial positions and velocities of the objects, the acceleration of the objects, and any external forces acting on the objects.

## What are some real-life examples of "catch-up" problems in 1-D kinematics?

Some real-life examples of "catch-up" problems in 1-D kinematics include a person running to catch a bus, a car merging onto a highway, or a boat catching up to another boat in a race.

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