Motion in One Dimension

  • #1

Homework Statement


#1
Runner A is initially 4.0 mi west of a flagpole and is running with a constant velocity of 6.0 mi/h due to east. Runner B is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when their paths cross ?

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#2
The engine of a model rocket accelerates the rocket vertically upward for 2.0s as folows: at t = 0, its speed is zero; at t = 1.0s, its speed is 5.0 m/s; at t = 2.0s, its speed is 16 m/s. Plot a velocity-time graph for this motion, and from it determine (a) the average acceleration during the 2.0 s interval and (b) the instantaneous acceleration at t = 1.5s.
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#3
In Mostar, Bosnia, the ultimate test of a young man's courage once was to jump off a 400year old bride (now destroyed) into the River Neretva, 23 m below the bridge. (a) How long did the jump last ? (b) How fast was the diver traveling upon impact with the river ? (c) If the speed of sound in air is 340 m/s, how long after the diver took off did a spectator on the bridge hear the splash ?
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#4
How can you find the average velocities and instantaneous velocities when you only look at the graph ? (use any graph that you want to show me)


Question ?
Those are 3 problems and 1 question that I need help on.
If you guys answer those problems, can you guys please show in details or show work for me (I'm not a smart person and i am really slow in understanding). And if possible , can you guys please analyze the problems for me ? I am really appreciate it. Thanks a lot guys ! I know that is a lot to you guys, but i have tried half my day on those and i couldn't figure it out.
 

Answers and Replies

  • #2
learningphysics
Homework Helper
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willingtolearn, it is against forum guidelines for us to just give the solution to you... can you show how you approached each problem? Where did you get stuck... what did you try etc... If you do that, we can try to help you along...
 
  • #3
willingtolearn, it is against forum guidelines for us to just give the solution to you... can you show how you approached each problem? Where did you get stuck... what did you try etc... If you do that, we can try to help you along...

oh, sorry about that ! I just joined the forum for 2 days.
 
  • #5
#1
From the problem we got:
Runner A : d=4 mi v = 6mi/h
Runner B : d =3 mi v=5mi/h, from that we have the distance from Runner A to Runner B = 7 mi and the velocity of two runner is 11 mi/h. To answer the question, first we have to find out when the Runner A and Runner B meet, (I stuck at this point, because i don't know how to calculate when they meet)

#2
From the problem we got:
a = -9.8 m/s2 (free falling object have an instant acceleration)
t = 0s, v = 0 m/s
t = 1s, v = 5 m/s
t = 2s, v = 16 m/s, and then i plot the t (s) in the x-axis, v (m/s) in the y-axis.
a) i got 8 m/s2( by divided 2 from 16, average acceleration = (final velocity - initial velocity)/time
***I only got to this point**

#3
From the problem we got
d = 23 m
a = -9.8 m/s2 (free falling object)
initial velocity = 0
final velocity = 0 (you start with 0, then you will end up with 0)


#4
Is that the same as finding the slope of a line ?
 
  • #6
learningphysics
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For part 1)

They are running towards each other... initial distance between them is 7mi. When they meet what can you say about the distances they both travelled? How does it relate to the 7mi? try to write expressions for distance in terms of t.

For part2)

8m/s^2 is right... for the instant acceleration, I think your teacher wants you to draw the tangent to the curve at t=1.5 and find the slope... ie that's the derivative of velocity at t=1.5, ie the acceleration at t=1.5

For part 3)

try to use the equation d = v1*t + (1/2)at^2, to get started...

For part 4)

Does he mean a position-time graph?
 
  • #7
#1
when they meet, the distance they both travelled different. d= rt

#2
I got it

#3
I got 5.3 sec for part a, but to part b i don't know what do

#4
Yes
 
  • #8
learningphysics
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#1
when they meet, the distance they both travelled different. d= rt

yes. one guy travels d = 6t. the other travels d = 5t. What can you say about the sum of the distances when they meet? Can you find the time they meet?

#2
I got it

#3
I got 5.3 sec for part a, but to part b i don't know what do

can you describe how you got 5.3? I get a different number. For part b, you just need to use v = v0 + at. where t is what you get from the first part.

#4
Yes

Instant velocity is just the slope of the tangent to the curve at any point on an x-t graph.

For average velocity from t1 to t2, draw the line from the point at t1, to the point at t2... the slope will be the average velocity: ie (x2-x1)/(t2-t1)
 
  • #9
#1
I get it. t = .64s

#3
I used v0t+(1/2)at^2=d

#4
I get it
 
  • #10
learningphysics
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#1
I get it. t = .64s

Cool. That's correct. Use that time to get that distance they meet from the flagpole

#3
I used v0t+(1/2)at^2=d

yes, that's right. I did the same thing but I don't get 5.3s... can you show your calculations?
 
  • #11
#3
d= -23 m (because he jumped down so it is a negative, is that right ?
v0t+(1/2)at^2=-23
(1/2) (-9.8)t^2=-23
-4.9t^2=-23
t= 2.17 s
My answer on previous post was wrong, i think i got a little mess up when i calculate them in my head.
 
  • #12
learningphysics
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4,099
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#3
d= -23 m (because he jumped down so it is a negative, is that right ?
v0t+(1/2)at^2=-23
(1/2) (-9.8)t^2=-23
-4.9t^2=-23
t= 2.17 s
My answer on previous post was wrong, i think i got a little mess up when i calculate them in my head.

Cool, that's right. How about part b) and c) for #3?
 
  • #13
#3
b) v = v0 + at
v = (-9.8)(2.17)
c) s = 340 m/s
d = 23 m and then we have t =.067 s
Is that right, i just guessing ...
 
  • #14
learningphysics
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#3
b) v = v0 + at
v = (-9.8)(2.17)

Looks right to me.

c) s = 340 m/s
d = 23 m and then we have t =.067 s
Is that right, i just guessing ...

That's the time from the moment the splash happened... but they're asking for the time from the moment she jumped... so you need to add 2.17s to 0.067s.

In your first post you said you were slow at understanding, but you don't seem slow at all. You were able to get through these problems well. :smile:
 
  • #15
Yeah! I understand them now !
Code:
In your first post you said you were slow at understanding, but you don't seem slow at all. You were able to get through these problems well.
Are you sure about that, I am the stupidest in my class. When the teacher did some lectures in class, they understood them all excepted me.
Anyway! Thanks a lot ... you are the man
 
  • #16
learningphysics
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6
Yeah! I understand them now !
Code:
In your first post you said you were slow at understanding, but you don't seem slow at all. You were able to get through these problems well.
Are you sure about that, I am the stupidest in my class. When the teacher did some lectures in class, they understood them all excepted me.

I bet a lot of them didn't understand. They just won't admit it. :wink: Just keep working hard, and you'll be fine.
 
  • #17
Thanks for your advice
 

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