Kinematics in 1 dimension


by future_vet
Tags: dimension, kinematics
future_vet
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#1
Jan30-07, 02:44 PM
P: 169
Hello! Could you please check for errors in the following HW problems?
Thank you!

A. 1. The problem statement, all variables and given/known data
A rolling ball moves from x1=3.4 cm to x2=-4.2 cm during the time from t1=3.0 s to t2=6.1 s. What is the average velocity?

2. Relevant equations and 3. The attempt at a solution
Average velocity = (x2 - x1)/(t2-t1) ~ (-2.5 cm/s)

B. 1. The problem statement, all variables and given/known data
A sports car accelerates from rest to 95 km/h in 6.2 s. What is the average acceleration in m/s^2?

2. Relevant equations and 3. The attempt at a solution
average acceleration = (v2 - v1)/(t2-t1) ~4.3 m/s^2.

C. 1. The problem statement, all variables and given/known data
At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s^2. At this rate, how long does it take to accelerate from 80km/h to 110 km/h?

2. Relevant equations and 3. The attempt at a solution
v1= 80 km/h = 22 m/s
v2= 110 km/h = 31 m/s
Average acceleration = 1.6m/s^2=(v2 - v1)/(∆t)=> it takes 5.6 seconds.

D. 1. The problem statement, all variables and given/known data
A horse canters away from its traiener in a straight line, moving 116m away in 14.0 seconds. It then turns abruptly and gallops halfway back in 4.8 seconds. Calculate its average speed and its average velocity for the entire trip, using away from the trainer as the positive direction.

2. Relevant equations and 3. The attempt at a solution
Average speed=d/t=(116 + 58)/(14.0 + 4.8 seconds) = 9.3 m/s.
Average velocity= ∆x/∆t=3.1m/s.

E. 1. The problem statement, all variables and given/known data
A light plane nust reach a speed of 33m/s for takeoff. How long a runway is needed if the constant acceleration is 3.0 m/s^2?

2. Relevant equations and 3. The attempt at a solution
(x-xo) = (v^2 -vo^2)/2a = (33m/s)^2/6.0= 181 meters.

F. 1. The problem statement, all variables and given/known data
In coming to a stop, a car leaves skid marks 92 m long on the highway. Assuming a decelration of 7.00m/s^2, estimate the speed of the car just before braking.

2. Relevant equations and 3. The attempt at a solution
V= √(2x7.00x92) ~ 36 m/s.

G. 1. The problem statement, all variables and given/known data
Estimate how long it took King Kong to fall straight down from the tip of the Empire State building (380 m high) and his velocity just before landing.

2. Relevant equations and 3. The attempt at a solution
y=vot1 + 1/2 x gt
t= 8.8 seconds.

V^2=Vo^2 + 2gh
V = 86 meters/seconds.

H. 1. The problem statement, all variables and given/known data
A ballplayer catches a ball 3.0 seconds after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

2. Relevant equations and 3. The attempt at a solution
I don't know which equation to use here... Could you please tell me which one it is?

Thanks!
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drpizza
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#2
Jan30-07, 03:20 PM
P: 291
Correct equation for B (if that's how your teacher gives you the equations)... How'd you get that answer? (A seems correct to me.)
Ditto for C. You have everything you need, including an equation that would work, but I can't figure out where you're getting that answer.
drpizza
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#3
Jan30-07, 03:30 PM
P: 291
DEFG are all fine.

Hint for H: when he throws the ball, it's going up for half the time, and coming down for half the time. Split the motion in half and look at either half.

cristo
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#4
Jan30-07, 03:31 PM
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Kinematics in 1 dimension


Quote Quote by future_vet View Post
Hello! Could you please check for errors in the following HW problems?
Thank you!

A. 1. The problem statement, all variables and given/known data
A rolling ball moves from x1=3.4 cm to x2=-4.2 cm during the time from t1=3.0 s to t2=6.1 s. What is the average velocity?

2. Relevant equations and 3. The attempt at a solution
Average velocity = (x2 - x1)/(t2-t1) ~ (-2.5 cm/s)
Correct

B. 1. The problem statement, all variables and given/known data
A sports car accelerates from rest to 95 km/h in 6.2 s. What is the average acceleration in m/s^2?

2. Relevant equations and 3. The attempt at a solution
average acceleration = (v2 - v1)/(t2-t1) ~4.3 m/s^2.
edit: yup, this is right; I didn't notice your change of units either!
C. 1. The problem statement, all variables and given/known data
At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s^2. At this rate, how long does it take to accelerate from 80km/h to 110 km/h?

2. Relevant equations and 3. The attempt at a solution
v1= 80 km/h = 22 m/s
v2= 110 km/h = 31 m/s
Average acceleration = 1.6m/s^2=(v2 - v1)/(∆t)=> it takes 5.6 seconds.
Correct

edit: late!
drpizza
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#5
Jan30-07, 03:34 PM
P: 291
edit: oh, never mind... carelessness on my part.
I wasn't paying careful enough attention to your units; shame on me!
(km/hr vs. m/s)
You were correct.

(You caught both of us off guard :) Both of the answers I said I didn't see where you got the solution from were in fact correct.) I rushed through your calculations a little too quickly.
future_vet
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#6
Jan30-07, 04:03 PM
P: 169
Hi!

So everything is correct up to G then? Thanks for your help :)

Let's see for H:
The ball is going to reach the apex after 1.5 seconds, where the velocity will be 0.
a=9.8 m/s^2, but will be negative on the way down...
We don't have the velocity
We don't have the height
We don't have the speed... but we know that the speed will be zero at the apex.

I have to say I'm still lost... It doesn't seem complicated at first, but I am not sure what to do here...

=/
cristo
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#7
Jan30-07, 04:16 PM
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Quote Quote by future_vet View Post
Hi!

So everything is correct up to G then? Thanks for your help :)

Let's see for H:
The ball is going to reach the apex after 1.5 seconds, where the velocity will be 0.
a=9.8 m/s^2, but will be negative on the way down...
We don't have the velocity
We don't have the height
You can calculate the height (at least the height relative to the point at which he throws the ball upwards, which I assume is what is required) but considering the point at which v=0 (hint: kinematic equations; more specifically, the one involving d, v, t and a).

Now you have h, you can calculate v.
We don't have the speed... but we know that the speed will be zero at the apex.
Speed is magnitude of velocity. Since the ball is thrown vertically, the velocity is equal to the speed here.
future_vet
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#8
Jan30-07, 04:29 PM
P: 169
d = v1 x t + 1/2 x a x t^2
d= 0 x 1.5 seconds + 1/2 x 9.8 x (1.5)^2
d= 11 meters.

Would this be correct for the height?

Thanks..
cristo
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#9
Jan30-07, 04:31 PM
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Quote Quote by future_vet View Post
d = v1 x t + 1/2 x a x t^2
d= 0 x 1.5 seconds + 1/2 x 9.8 x (1.5)^2
d= 11 meters.

Would this be correct for the height?

Thanks..
That's correct, although your original equation should read d=vt-1/2at2. a=-9.8, and so this becomes your second line.
future_vet
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#10
Jan31-07, 04:11 PM
P: 169
For the height:
d=0x1.5 + 1/2 x (-9.8) x (1.5)^2 then?

From there, I should have all I need to find the speed?

I had no problem doing the homework up to this point, but for some reason it's very confusing to me =/

Thanks!
cristo
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#11
Jan31-07, 04:28 PM
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Quote Quote by future_vet View Post
For the height:
d=0x1.5 + 1/2 x (-9.8) x (1.5)^2 then?

From there, I should have all I need to find the speed?

I had no problem doing the homework up to this point, but for some reason it's very confusing to me =/

Thanks!
you have calculated the height above (using d=vt-1/2at2). Now, you can use a similar equation for initial velocity, and consider the whole journey of the ball: d=ut+1/2at2. You know the time for the whole flight, a, and d is the value you have just calculated, so you can solve for u (the initial velocity)
future_vet
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#12
Jan31-07, 04:36 PM
P: 169
Wit a= -9.8, I get a negative height, and then a negative speed...

Here's what I did:
d=vt-1/2at^2
d= 0 x 1.5 + 0.5 x 9.8 x 1.5^2 = - 11 meters

For the speed:
d (above) = vt-1/2at^2 = v x 3 seconds - 0.5 x (-9.8) x 3^2

What am I doing wrong?...

Thank you and sorry for all the trouble :)
cristo
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#13
Jan31-07, 04:41 PM
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Quote Quote by future_vet View Post
Wit a= -9.8, I get a negative height, and then a negative speed...

Here's what I did:
d=vt-1/2at^2
d= 0 x 1.5 + 0.5 x 9.8 x 1.5^2 = - 11 meters
This isn't correct. This d= 0 x 1.5 + 0.5 x 9.8 x 1.5^2 is correct, but it does not equal -11; it is equal to 11!

Note that the original equation has the term -1/2at^2 in it, but when you substitute in a=-9.8, this term becomes + 0.5 x 9.8 x 1.5^2 (like you did above). The correct answer is 11m.

For the initial speed, you must use the equation d=ut+1/2at^2 (note the difference in sign between the equation for initial, and final, velocity). Sub in what you know and you should be fine.
future_vet
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#14
Jan31-07, 04:53 PM
P: 169
So, I think I got it now..

For the height, you were right, I apologize for that!

For the speed:
11 = 3v + 1/2 x -9.8 x 3^2 (I use the whole 3 seconds for the trajectory, not half the seconds, right?)
3v = 55.1
v= 18 meters per second.

Good?
cristo
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#15
Jan31-07, 07:18 PM
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Quote Quote by future_vet View Post
So, I think I got it now..

For the height, you were right, I apologize for that!
That's ok; it's more important that you recognise the difference between the two equations, than you make a simple algebraic mistake.

For the speed:
11 = 3v + 1/2 x -9.8 x 3^2 (I use the whole 3 seconds for the trajectory, not half the seconds, right?)
3v = 55.1
v= 18 meters per second.

Good?
Yup, that's correct.
future_vet
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#16
Jan31-07, 07:22 PM
P: 169
Thank you!!


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