# Questions in linear motion and rotating motion

1. May 24, 2012

### manal950

Hi

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

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

Last edited by a moderator: May 6, 2017
2. May 24, 2012

### Fewmet

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.

3. May 25, 2012

### manal950

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

4. May 25, 2012

### HallsofIvy

Staff Emeritus
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.

5. May 25, 2012

### azizlwl

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

6. May 25, 2012

### 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 accleration = a = -6.6

7. May 25, 2012

### Fewmet

Correct except the answer rounds to -6.7 m/s2
Correct.

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.

8. May 25, 2012

### manal950

thank sou much what about second one ...

9. May 25, 2012

### Fewmet

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...

10. May 25, 2012

### 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.

11. May 26, 2012

### 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 ?

12. May 26, 2012

### azizlwl

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

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

Last edited: May 26, 2012
13. May 26, 2012

### 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
but now we have 2 variable a and w0

How I can get one of them ????

14. May 26, 2012

### 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

Last edited: May 26, 2012
15. May 26, 2012

### manal950

16. May 26, 2012

### 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 ??

17. May 26, 2012

### 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

Last edited: May 26, 2012
18. May 26, 2012

### manal950

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

w = w0 + at
a = (12.5 - 20 ) 8
= -0.935

is correct ..?

19. May 26, 2012

### azizlwl

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

Now we are talking on acceleration NOT equal to zero.

20. May 26, 2012

### manal950

12.5 - 20 / 18

a =-0.4

is correct ?

give me the way to find a ? ??????????????????????????????????????????????????????