Questions in linear motion and rotating motion

[manal950]

Hi

please I want from you check my answer

http://store2.up-00.com/May12/Ueu86896.jpg

http://store2.up-00.com/May12/fkl86896.jpg

 

[Fewmet]

Hi

please I want from you check my answer
[/PLAIN]

In this case the equation
V²=u²+2as
means that the square of the final velocity equal the square of the initial velocity plus two times the stopping acceleration time the distance over which the car stopped. Think through the question and see that the car does not accelerate over a distance of 60 m.

If you substitute the correct distance, you'll get the correct answer.

 

[manal950]

But how I can know the total distance where the car stop

 

[HallsofIvy]

Yes, you do. You are told that the driver sees a person 60 m ahead of him. You are also told the he continues at 20m/s for 1.5 s before braking. At 20 m/s he will go (20)(1.5)= 30 m in 1.5 seconds. He has the remaining 60- 30= 30 m in which to stop.

 

[azizlwl]

You are using the same variables, s and t, for given values and required values.
You should consider what is meant by delay.

 

[manal950]

1 )

u = 20
V = 0
s = 30 m

a ) V^2 = u^2 + 2as
0 = 20^2 + 2 X a X 30
a = -6.6

time requried

V = u + at
t = 3 s

now distance

s = ut + 1/2 a t^2
s = 20 X(3) + 1/2 X(-6.6) X 3^2
s = 30 m

uniform acceleration = a = -6.6

 

[Fewmet]

1 )

u = 20
V = 0
s = 30 m

a ) V^2 = u^2 + 2as
0 = 20^2 + 2 X a X 30
a = -6.6

Correct except the answer rounds to -6.7 m/s²

time requried

V = u + at
t = 3 s

Correct.

now distance

s = ut + 1/2 a t^2
s = 20 X(3) + 1/2 X(-6.6) X 3^2
s = 30 m

uniform acceleration = a = -6.6

Yes, but you have already calculated ( or, rather, HallsofIvy calculated) how far the car takes to stop. You did not need to do it again.

 

[manal950]

thank sou much what about second one ...

 

[Fewmet]

thank sou much what about second one ...

I am wrestling with the wording a little...You have correctly calculated the average rotational velocity as 20 rad/s for the first 10 seconds. The given information says the next eight seconds has a lower average rotational speed. In other words, there was a negative acceleration, so your answer that angular acceleration is zero is not correct.

What bugs me here is that you can find an average rotational velocity from t=10 s to
t =18 s, but acceleration should be calculated from instantaneous velocities. I am guess you are expected to figure that the average rotational velocity occurs at the center of each interval...

 

[azizlwl]

Conceptualize.
If a diagram is not provided, you should almost always make a quick drawing of the situation.
Shouldn't rely on equations for plug and chug.

For a constant acceleration kinematic, you should be able to draw and interpret a velocity-time diagram.

 

[manal950]

Can I do by this
Q = w0t + 1/2 at^2
200 = 10W0 + 50a ====== first equation Q = w0t + 1/2 at^2
100 =8w0 + 32a ======== second equation

200 = 10W0 + 50a
100 =8w0 + 32a
-------------------------
100 = 2W0 + 18a is this correct until now ?

 

[azizlwl]

Segment #2
Constant velocity->8 secs -> 100reve.
Constant velocity?
Equation?

Segment #1
Constant acceleration ->10 secs -> 200 rev.
Equation?

 

[manal950]

see the Q2 ... It want initial velocity during the acceleration was constant and after change to the a = 0 as the velocity is constant

now I will ask is the initial velocity depend for constant acceleration or from it was constant to changed .

if the initial velocity depend only to first rev that mean
w = w0 + at
w = 20 rad/s
but now we have 2 variable a and w0

How I can get one of them ?

 

[azizlwl]

What is the value of constant velocity?

This value is the v of constant acceleration.

s=ut+1/2at²
a=(v-u)/t

s=ut+1/2t²(v-u)/t

2s=2ut + vt-ut

u=(2s-vt)/t

 

[manal950]

constant velocity is 12.5 rad/s

 

[manal950]

OMG ! from where you got this equations

We learn only 3 equation
w = w0 + at
w2 = w0^2 + 2sQ
Q = W0t + 1/2 at^2

I am confusing ??

 

[azizlwl]

These 2 are standard.
s=ut+1/2at2 ...(1)
a=(v-u)/t ...(2)

Now you substitute "a" in equation (1)
And you get
s=ut+1/2t2(v-u)/t

Just replace the translational variables with rotational variables.
u ->ω₀
v ->ω₁
a ->α
s->θ

 

[manal950]

why substitute ... why we don't take value for each variables ..

w = w0 + at
a = (12.5 - 20 ) 8
= -0.935is correct ..?

 

[azizlwl]

ω₁=12.5 -Correct.
Where do get the value ω₀=20?

Now we are talking on acceleration NOT equal to zero.

 

[manal950]

12.5 - 20 / 18

a =-0.4

is correct ?

give me the way to find a ? ??

 

[manal950]

is the final answer for a = - 1.5 and velocity = 2.7 ?

 

[azizlwl]

Graphical representation might help you to understand.
Note the object state from 1->2->3

http://img220.imageshack.us/img220/8412/p1010010je.jpg

 

[manal950]

But is the final answer for a = - 1.5 rad/s^2 and velocity = 2.7 rad/s?

 

[rl.bhat]

But is the final answer for a = - 1.5 rad/s^2 and velocity = 2.7 rad/s?

Acceleration is correct. Check the velocity.

 

[manal950]

ok I got the answer 27... thaaaaaaaank.. can help my to draw for this
https://www.physicsforums.com/showthread ... 611&page=2

and it same way to solve that Q and this one ...

and can give me all steps to solve like this Q (In fact I face only problem in like this Q when the acceleration change ex was constant change to zero )

 

[Fewmet]

can help my to draw for this
https://www.physicsforums.com/showthread ... 611&page=2

That link is not opening for me.