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Kinematics Equation Problem!

  • Thread starter harujina
  • Start date
  • #1
77
1

Homework Statement



Sam is running at 3.8 m/s and is 75m behind John who is running at a constant velocity of 4.2m/s. If Sam accelerates at 0.15m/s^2, how long will it take him to catch John?

Homework Equations



d = (vf+vi/2)t
vf = vi + at
d = vit + 1/2at^2
vf^2 = vi^2 + 2ad
d = vft - 1/2at^2


The Attempt at a Solution



I have asked this before, but have now come back with a little more understanding than before.
I've done the following so far:

Xs (displacement of Sam) = viΔt + 1/2aΔt^2
= (3.8m/s)Δt + (0.075m/s^2)Δt^2

Xj (displacement of John) = 75 + 4.2m/sΔt

[-> from: (xf-xi) = VavΔt
xf = xi + VavΔt ]

---

... Ok so now I do (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt
to find the time when they are in the same position/displacement, correct?

But I'm so confused as to what I would do after that?

EDIT; Also, I got 1005.33s for the time, and I don't think that's right?
 
Last edited:

Answers and Replies

  • #2
6,054
390
What you got is a quadratic equation where delta t is unknown. Solve it.
 
  • #3
77
1
ax^2 + bx + c = 0?

This may be a stupid question but how would I know which variable goes for which coefficient?
(I need to brush up on my math...)
 
  • #4
6,054
390
ax^2 + bx + c = 0?

This may be a stupid question but how would I know which variable goes for which coefficient?
(I need to brush up on my math...)
##x## is the unknown. In your case, the unknown is ##\Delta t##. You should be able to match the other symbols.
 
  • #5
77
1
Okay I think I get it but then what was this for: (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt ?
I thought I was trying to find time with that?
 
  • #6
6,054
390
I do not understand your questions. What does ##\Delta t## mean to you?
 
  • #7
77
1
time, is it not?
 
  • #8
6,054
390
It is time, and it is what you are after. What is not clear to you?
 
  • #9
77
1
Well, my teacher said I had to find the position/displacement of both Sam and John, which I did:
(3.8m/s)Δt + (0.075m/s^2)Δt^2 for Sam and
75 + 4.2m/sΔt for John

and if I do: (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt
won't that give me time? So why would I have to do the quadratics equation?
Or am I doing something wrong here...
 
  • #10
6,054
390
and if I do: (3.8m/s)Δt + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt
won't that give me time? So why would I have to do the quadratics equation?
Because the equation you got is a quadratic equation. It will give you time - when you solve it.
 
  • #11
77
1
OH, now I understand what my teacher was saying!
I got 32s, is that right?
 
  • #12
6,054
390
You can check your answer yourself: you have the formula for displacement for Sam, and the formula for John. Do you get the same displacements at 32 s?
 
  • #13
31
0
I got 34.4 seconds
 
  • #14
77
1
(3.8m/s)(32s) + (0.075m/s^2)(32s)^2 = 198.4 m
75 + 4.2m/s(32s) = 209.4 m

so I am a bit off...
this is what i did:
---
(3.8m/s)Δt/t + (0.075m/s^2)Δt^2 = 75 + 4.2m/sΔt/t
3.8m/s - 3.8m/s + (0.075m/s^2)Δt^2 = 75 + 4.2m/s - 3.8m/s
(0.075m/s^2)Δt^2 / 0.075m/s^2 = 75 + 4.2m/s - 3.8m/s / 0.075m/s^2
Δt^2 = 75 + 4.2m/s - 3.8m/s / 0.075m/s^2
(square root)Δt = (square root)1005.33 s

Δt = 32 s
 
Last edited:
  • #15
6,054
390
I do not know what you are doing, but you are not solving the equation correctly.

First you need to bring it to the standard form: ## a \Delta t^2 + b \Delta t + c = 0 ##.

Then use the (supposedly) well-known formula for its roots.
 
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  • #16
77
1
Okay I finally got it. Thank you so much and sorry for all the troubles!
 
  • #17
6,054
390
No trouble at all. I hope this has been useful for you :)
 

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