Acceleration Equation - Two Objects Moving Forward; Meeting Point?

1. Mar 3, 2013

zak.hja

1. The problem statement, all variables and given/known data
A child stands 30m behind an ice-cream truck. The child then runs towards the truck at a constant velocity of 5m/s. At that moment, the ice-cream truck accelerates at 1m/s^2 from rest to drive away from the child. Does the child catch the ice-cream truck?

2. Relevant equations
Δd=v(initial)t-1/2∂Δt^2
d=(vf+vi)/2*t
vf=vi+at
d=(vi)t+1/2at^2
d=(vf)t-1/2at^2
v=Δd/Δt
∂=v/Δt

3. The attempt at a solution
Bearing in mind that I had to solve for time, I made the assumption that if the two objects met, their times would be the same (t1=t2). However, I'm not sure how to approach this question and I'm very new to these types of problems; any help is appreciated.

Last edited: Mar 3, 2013
2. Mar 3, 2013

cepheid

Staff Emeritus
Welcome to PF zak.hja,

Sure, it goes without saying that they meet up at a particular instant in time. What is more important, is that if they meet up, then their positions are the same (i.e. they are in the same place). Can you write down an expression for the position vs. time of each of them (the child and the ice cream truck)?

3. Mar 3, 2013

zak.hja

Thank-you :)
So wait, which expression could I use with position vs. time? These are the equations that I have learned so far:

d=(vf+vi)/2*t
vf=vi+at
d=(vi)t+1/2at^2
d=(vf)t-1/2at^2

4. Mar 3, 2013

cepheid

Staff Emeritus
The kid is accelerating at a constant rate. So you'd used the distance vs. time equation that applies for the case of constant acceleration. (So, clearly, it should have an "a" in it).

The ice cream truck is not accelerating. If the acceleration is zero, the speed is constant. How is distance related to speed and time when the speed is constant? This is the simplest equation out of all of them.

5. Mar 4, 2013

zak.hja

So velocity is equal to distance divided by time, or in the case of the accelerating ice cream truck, acceleration is equal to change in velocity divided by the time.

Hang on, the child is running 30m initially before he even reaches the same point at which the truck began accelerating. So after 6 seconds (30m/5m/s), he reaches the initial starting point of the truck. And using the a=Δv/t equation with the 6 seconds and the 1m/s^2 acceleration, the truck's Δv would be 6m/s. What should be solved next?

6. Feb 16, 2015

tycoon515

7. Feb 16, 2015

BvU

Dear zak, welcome to PF once more

If you list off a load of equations like you did, I'm afraid they don't mean much to you.

So my first advice is: familiarize yourself with uniform linear motion, find out what the customary variable names mean (x, v, s, t). Then idem with uniformly accelerated linear motion. There are 6 formulas here and if you grasp numbers 4 and 5 you're well equipped to deal with most of the exercises -- provided your math is OK.

Second advice it to make a sketch: time horizontal, position vertical.
Kid is at x=0m for t=0s and runs 5m/s. Uniform motion, constant speed. So that's a straight line with slope 5 m/s. As you already found out, 30 m takes 6 s.
Truck is at 30 m for t=0s and accelerates with 1 m/s2. For that you need the equation for uniformly accelerated linear motion (number 5). It starts, so v0 = 0 m/s. And x0 = 30 m. A parabola. x = 31 m at t=1s, x = 34m at t=2s, 39m at t=3s etc.

Third advice is to brush up on parabola equations (most of these exercises have to do with parabolas, so you want to be able to solve quadratic equations). But quite often a simplification/short is available for smart or lazy (or both) students.

In this case there's a smart way out too, and in fact you have already found it (in post #5, but do you realize you did ?) : by the time the kid is at 30 m, the truck speed is > kid speed, so he can forget about ever catching up. Nothing to solve. Wait for the next ice truck is all the poor bugger can do

8. Feb 17, 2015

tycoon515

Pay special attention to the emboldened details. We're told that the child starts behind the truck and is running towards it at constant velocity. Meanwhile, the truck starts at rest ahead of the child and begins accelerating away from the child. Therefore, the child and the truck are both headed in the same direction. If we let x1(t) and x2(t) represent the positions of the child and the truck, respectively, and we define the initial position of the child to be 0 m and the direction the child is running to be the positive direction, then we can say the truck's initial position is 30 m, its initial velocity is 0 m/s, and its acceleration is 1 m/s^2 while the child's initial velocity is 5 m/s, which is also his or her velocity at any other point in time. In order to answer the question, observe that after 5 seconds, the child's position is 25 m, still 5 m behind where the truck started, but now the truck's position is farther in the positive direction than where it started, and it has already attained a speed of 5 m/s, the same speed as the child runs, and it will continue to speed up, so it seems reasonable that the child never reaches the truck. It is easy to prove that the closest the child gets to the truck is at 5 seconds because before 5 seconds, the truck is traveling slower than 5 m/s, and after 5 seconds, the truck is traveling faster than 5 m/s. Hopefully this intuitive approach helps.