JSC_23
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Physics Help!

Sun Mar 16, 2003 6:29 pm

for homework we have this question :
"while travelling at 40 ms a driver sees a red light 100m away.Calculate the steady acceleration he must give the car if he is to stop at the light"...

i can't work out the formula to use...(or how to rearrange one..)basically the things we are given are
velocity initial:40ms
velocity final:0 ms
distance:100m

and i need to find accelleration?

any ideas....help would be much appreciated...

kindest regards
JSC_23
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POSITIVE RATE
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RE: Physics Help!

Sun Mar 16, 2003 7:11 pm

This is very rudimentary in my head calculations here so someone please correct me if i'm wrong. If you're doing 40m/s(144km/h) and you see a red light 100m away, you will need to decelerate at a steady 16m/s in order to stop in time.
 
JSC_23
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RE: Physics Help!

Sun Mar 16, 2003 7:13 pm

Im sorry for this but also I have this question:
"how long does it take to hit the ground if you jump off a wall 3m high?"
I work out that the acceleration is 10 m/s/s (gravity) so it should work out to about 1/3rd of a second 0.333333333 but it seems to easy?

JSC_23
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fritzi
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RE: Physics Help!

Sun Mar 16, 2003 7:15 pm

Thake the equation:

d=((Vi+Vf)/2)*T

rewrite it so that you solve for T

T=(2/(Vi+Vf))*d

T=(2/(40+0))*100

T=(2/(40))*100

T=(0,05)*100

T= 5 m/sec(sq)
 
fritzi
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RE: Physics Help!

Sun Mar 16, 2003 7:19 pm


T= - 5 m/sec(sq)
 
JSC_23
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RE: Physics Help!

Sun Mar 16, 2003 7:24 pm

Thanks Fritzi
-Please excuse me as I'm a beginner but isn't that working out time and not acceleration????

JSC_23
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fritzi
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RE: Physics Help!

Sun Mar 16, 2003 7:28 pm

q2)

d=0,5GT(squared)

3=0,5*-10T(Sq)

3/(5) = T(Sq)

(Sq root)0,6= (Sq root)T(Sq)
T =0,775 sec

I did this one very quickly so there may be some mistakes
 
fritzi
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RE: Physics Help!

Sun Mar 16, 2003 7:30 pm

The answer for acceleration is positive, the answer for decelleration is negative.

Its the same formula.
 
fritzi
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RE: Physics Help!

Sun Mar 16, 2003 7:33 pm

I may have screwed up but I dont have any more time right now because I have my IB A1 HL Swedish oral in 1.5 hours and I am a bit stressed
 
JSC_23
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RE: Physics Help!

Sun Mar 16, 2003 7:34 pm

im sure your right but i can't understand what the T is for in the first equation....shouldn't it be A=,???

Thanks for you help...

JSC_23
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fritzi
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RE: Physics Help!

Sun Mar 16, 2003 7:36 pm

Vf(Sq)=Vi(Sq)+(2*A*2) may be the equation that you are looking for.

Sorry if I mislead you. As I said, I am under a lot of pressure right now.
 
Sunair
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RE: Physics Help!

Sun Mar 16, 2003 7:43 pm

Uhh, or you just use this equation:

v=u + 2as

v=0ms
u=40ms
s=100m

0=40 + 2(a)(100)
-40 = 200a
a=-5ms^-2
Therefore acceleration = 5ms^-2 deceleration
 
JSC_23
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RE: Physics Help!

Sun Mar 16, 2003 7:43 pm

Please do NOT be sorry........

You are helping me alot....i very much appreciate that!!!!!

JSC_23
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JSC_23
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RE: Physics Help!

Sun Mar 16, 2003 7:52 pm

BEST OF LUCK
for your exam Fritzi...i hope everything goes well!!!!!!!!!

JSC_23
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delta-flyer
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RE: Physics Help!

Mon Mar 17, 2003 12:03 am

Sorry, guys, all wrong!!! Let an old fart do it right......

Start with the distance-time-acceleration formula:

d = 0.5 * a * t^2 ...EQ(1)

d = distance = 100m
a = acceleration = ?
t = time to decelarate from 40 m/s to 0 = ? ( t^2 = t "squared")

Next, the definition of acceleration is rate of change of speed ..... at constant acceleration, it is simply the change in speed divided by the time taken for that change .....

a = (Vf - V0) / t ...EQ(2)

Vf = final velovity = 0 m/s
V0 = initial velocity = 40 m/s (by the way, velocity is m/s, not ms!)

Note that we have velocities and distance, want to solve for acceleration. So let's eliminate time from equation (1) by substituting equation (2), thus ....

Rewriting EQ(2):

t = (Vf - V0) / a ...EQ(3)

Substitute for t in EQ(1):

d = 0.5 * a * { (Vf V0) / a }^2

Collect terms, simplify:

d = 0.5 * (Vf - V0)^2 / a or,

d = (Vf - V0)^2 / (2 * a)

Now, rearrange to solve for a:

a = (Vf V0)^2 / (2 * d)

Finally, plug in the numbers:

a = (0 - 40)^2 / (2 * 100)
= 1600 / 200
= 8 m/s^2 (metres per second per second)

Now, to check:
Substitute into equation (3):
t = (0 - 40) / 8 = 5 seconds

Substitute into equation (1), using t = 5 s:
t = 0.5 * 8 * (5)^2
= 0.5 * 8 * 25
= 100 metres

It checks out!!
=====================

Now, Fritzi was almost there .... he solved for time correctly (5 sec) but inadvertantly called it acceleration ... see above ... he gave T = 5 m/s (sq) should have been 5 sec.

======================

Here's a trick: you can easily determine if an equation is NOT correct if the units don't work out. For example, SunAir gave v=u + 2as (v, u are velocity, a is acceleration, s is distance) - here's how to check it out .....

Substitute units for variables:

v = u + 2as
m/s = m/s + (2) m/s/s * m {note the constant 2 has no units - just ignore it}

Note the last term, when you simplify, becomes m*m /s/s or (m/s)^2
That is velocity squared, and is inconsistent with the other terms which are velocity.

This methodology is a part of "dimensional analysis", which is used in some sciences to construct equations where physical models are not well understood.

=================

You second problem can also use equation (1), but you have to rewrite it as a function of time:

t^2 = d / (0.5 * a) but a = accel of gravity, or g = 10 m/s^2

t^2 = d / (0.5 * g)

t = sqrt { d / (0.5 * g) }

Plug in numbers ......

t = sqrt { 3 / (5) } = sqrt { 0.6 } = 0.775 seconds

Fritzi is correct!!!

I hope you can follow this. Please don't just copy it, make sure you understand it. I'll be glad to answer any questions. I'm working on my taxes today, so I'll be on my computer off and on all day.

Good luck,
Pete









"In God we trust, everyone else bring data"
 
Sunair
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RE: Physics Help!

Mon Mar 17, 2003 3:55 pm

"This methodology is a part of "dimensional analysis", which is used in some sciences to construct equations where physical models are not well understood."

So are these equations (v=u+at, v^2=U^2+2as etc...) incorrect? These equations are part of what I got my final physics school certificate on!  Wow!
 
delta-flyer
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RE: Physics Help!

Fri Mar 21, 2003 12:19 pm

So are these equations (v=u+at, v^2=U^2+2as etc...) incorrect?

No, the above equations are quite correct. However, you wrote:

v=u + 2as


which is incorrect, and led you to the wrong answer. My dimensional analysis example also showed that this equation cannot be correct, as the units are inconsistent.

Sorry,
Pete
"In God we trust, everyone else bring data"
 
flight152
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RE: Physics Help!

Fri Mar 21, 2003 12:41 pm

Where is JetService when you need him?  Laugh out loud
 
delta-flyer
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RE: Physics Help!

Fri Mar 21, 2003 1:10 pm

This problem has obviously left him .......

..... speechless?!!

 Big grin
"In God we trust, everyone else bring data"
 
redngold
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RE: Physics Help!

Fri Mar 21, 2003 1:29 pm

NEGATIVE!

Remember, if you are decelerating your answer must be a negative number!


redngold
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