JSC_23
Topic Author
Posts: 214
Joined: Fri Jul 20, 2001 8:50 pm

### Physics Help!

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
Be a realist...Remain a dreamer!

POSITIVE RATE
Posts: 2121
Joined: Sun Sep 16, 2001 11:31 am

### RE: Physics Help!

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
Topic Author
Posts: 214
Joined: Fri Jul 20, 2001 8:50 pm

### RE: Physics Help!

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
Be a realist...Remain a dreamer!

fritzi
Posts: 2598
Joined: Sat Jun 30, 2001 2:34 am

### RE: Physics Help!

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
Posts: 2598
Joined: Sat Jun 30, 2001 2:34 am

### RE: Physics Help!

T= - 5 m/sec(sq)

JSC_23
Topic Author
Posts: 214
Joined: Fri Jul 20, 2001 8:50 pm

### RE: Physics Help!

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

JSC_23
Be a realist...Remain a dreamer!

fritzi
Posts: 2598
Joined: Sat Jun 30, 2001 2:34 am

### RE: Physics Help!

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
Posts: 2598
Joined: Sat Jun 30, 2001 2:34 am

### RE: Physics Help!

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

Its the same formula.

fritzi
Posts: 2598
Joined: Sat Jun 30, 2001 2:34 am

### RE: Physics Help!

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
Topic Author
Posts: 214
Joined: Fri Jul 20, 2001 8:50 pm

### RE: Physics Help!

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
Be a realist...Remain a dreamer!

fritzi
Posts: 2598
Joined: Sat Jun 30, 2001 2:34 am

### RE: Physics Help!

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
Posts: 346
Joined: Thu Jun 11, 2009 6:59 am

### RE: Physics Help!

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
Topic Author
Posts: 214
Joined: Fri Jul 20, 2001 8:50 pm

### RE: Physics Help!

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

JSC_23
Be a realist...Remain a dreamer!

JSC_23
Topic Author
Posts: 214
Joined: Fri Jul 20, 2001 8:50 pm

### RE: Physics Help!

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

JSC_23
Be a realist...Remain a dreamer!

delta-flyer
Posts: 2631
Joined: Mon Jul 30, 2001 9:47 am

### RE: Physics Help!

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

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
Posts: 346
Joined: Thu Jun 11, 2009 6:59 am

### RE: Physics Help!

"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!

delta-flyer
Posts: 2631
Joined: Mon Jul 30, 2001 9:47 am

### RE: Physics Help!

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
Posts: 3286
Joined: Fri Nov 24, 2000 8:04 am

### RE: Physics Help!

Where is JetService when you need him?

delta-flyer
Posts: 2631
Joined: Mon Jul 30, 2001 9:47 am

### RE: Physics Help!

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

..... speechless?!!

"In God we trust, everyone else bring data"

redngold
Posts: 6673
Joined: Wed Mar 22, 2000 12:26 pm

### RE: Physics Help!

NEGATIVE!

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

redngold
Up, up and away!

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