Need a little help with some homework

  • Thread starter tehmatriks
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Homework Statement


A body leaves a point A and moves in a straight line with a constant velocity of 40 m s-1. Ten seconds later another body which is at rest at A is given an acceleration of 2 m s-2 and moves in the same direction as the first body. How long does it take the second body to catch up with the first? How far from A does this occur?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
gneill
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You've got to make an attempt and show your work at least up to the point where you get "stuck" before help can be offered.

How do you think you might approach the problem? What equations of motion might be relevant?
 
  • #3
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sorry bout that gneill, forgot to put in the relevent equations:

Homework Equations


[v = u + at] [s = ut + ½at2] [v2 = u2 + 2as]


The Attempt at a Solution

[/QUOTE]

is it possible to get stuck right in the beggining? and you dont even know how to start?
here's my sad attempt

s = (40)(10) + ½(2)(10)2
s = 400 + 100
s = 500 m
 
Last edited:
  • #4
verty
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Hint: What is special about the point where they meet?
 
  • #6
gneill
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Suppose we call the bodies body 1 and body 2. Body 1 is the one that leaves first, moving at a constant 40m/s. Body 2 is holding a stopwatch which begins counting from zero as soon as body two starts its journey, accelerating at 2 m/s2.

The first question to ask is, how far apart are the two bodies at stopwatch t=0? Body 1 has had a ten second head start. So do = ?
 
  • #7
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Suppose we call the bodies body 1 and body 2. Body 1 is the one that leaves first, moving at a constant 40m/s. Body 2 is holding a stopwatch which begins counting from zero as soon as body two starts its journey, accelerating at 2 m/s2.

The first question to ask is, how far apart are the two bodies at stopwatch t=0? Body 1 has had a ten second head start. So do = ?

400m
 
  • #8
gneill
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Okay. So at t=0 the first body (let's call it B1) has a 400m head start. Can you write an equation of motion for body B1 from this t=0 time onwards?
 
  • #10
gneill
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It's moving at constant velocity v1 = 40m/s with an initial distance of do = 400m. So d = ???
 
  • #11
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It's moving at constant velocity v1 = 40m/s with an initial distance of do = 400m. So d = ???

what is do, i thought s= distance, also if do is distance what's the difference between do and d?

im also guessing in these types of questions we wont be using the 3 formulas: [v = u + at], [s = ut + ½at2], [v2 = u2 + 2as]

aw man, i think im gonna completely skip this question
 
  • #12
gneill
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Those formulas are quite applicable. But you should get used to using different variable names and seeing the same formulas expressed using different names. I didn't realize that you were used to seeing s as a distance (to me, s means the unit "second"). Velocities can be u, v, ... anything you wish as long as you're consistent.

So, if I were to rename the variables... the initial distance is so = 400m and the velocity is u = 40m/s. Write the expression for distance s with respect to time.
 
  • #13
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hey gneill, i hate to waste your time like this on someones as clueless as me, so lets try and wrap it up

just show me how you would go about doing it, your way, i'll learn from the whole thing, that's what i need, right now your killing me
 
  • #14
gneill
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hey gneill, i hate to waste your time like this on someones as clueless as me, so lets try and wrap it up

just show me how you would go about doing it, your way, i'll learn from the whole thing, that's what i need, right now your killing me

I'm very sorry, but forum rules prevent supplying the answer directly. I can only help you understand and lead you to find the solution, point out errors, and so on.

If my methods are not to your liking, perhaps someone else will step in and give it a go.

Again, sorry about that.
 
  • #15
rl.bhat
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hey gneill, i hate to waste your time like this on someones as clueless as me, so lets try and wrap it up

just show me how you would go about doing it, your way, i'll learn from the whole thing, that's what i need, right now your killing me
Hi tehmatriks, don,t loss your patience. Instead of giving the answer directly we are trying to teach the concepts involved in the problem, so that later on you can solve the similar problem your self. Now try this method.
Let Tom and Dick start from A and meet at B.
Tom start first. He moves with uniform velocity v. So the acceleration is zero. He takes time t to reach B. Using the proper kinematic equation, can you find s, the distance between A and B, in terms of s and t? Keep the unknown quantities as it is.
Dick starts from rest at A 10 s later with uniform acceleration a. He meets Tom at B after ( t - 10) s. Write down the equation for displacement s.
When they meet at B, both must have traveled the same distance s.
So equate the two equations and solve the quadratic to find t. You will get two values. Select the value which is more than 10 s.
 
  • #16
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oh....a = 0, *facepalm*

tell me if i'm on the right path

body1
[s = ut + ½at2]
s = 40t + ½(0)t2
s = 40t

and did you not mean (10 + t)?

body2
[s = ut + ½at2]
s = (0)t + ½2(10+t)2
s = (10 + t)2

and for the b formula
b = u, a = a, c = 10 ??

-40 +- sqr402 - 4(2)(10)/ 2(2)
-40 +- sqr1600 - 80/ 4
-40 +- sqr1520/4
-40 +- 38.987/4

ok....i've done it wrong

i have to get home and take a closer look at it
 

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