JSC_23 From New Zealand, joined Jul 2001, 223 posts, RR: 0 Posted (11 years 9 months 2 weeks 2 hours ago) and read 1087 times:

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

Positive rate From Australia, joined Sep 2001, 2143 posts, RR: 1
Reply 1, posted (11 years 9 months 2 weeks 2 hours ago) and read 1053 times:

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 From New Zealand, joined Jul 2001, 223 posts, RR: 0
Reply 2, posted (11 years 9 months 2 weeks 2 hours ago) and read 1052 times:

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?

Delta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 6
Reply 14, posted (11 years 9 months 1 week 6 days 21 hours ago) and read 966 times:

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

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.

SunAir From India, joined Jun 2009, 0 posts, RR: 1
Reply 15, posted (11 years 9 months 1 week 6 days 5 hours ago) and read 899 times:

"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 From United States of America, joined Jul 2001, 2676 posts, RR: 6
Reply 16, posted (11 years 9 months 1 week 2 days 9 hours ago) and read 849 times:

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.

Flight152 From United States of America, joined Nov 2000, 3413 posts, RR: 6
Reply 17, posted (11 years 9 months 1 week 2 days 8 hours ago) and read 846 times:

Delta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 6
Reply 18, posted (11 years 9 months 1 week 2 days 8 hours ago) and read 836 times:

Redngold From United States of America, joined Mar 2000, 6907 posts, RR: 44
Reply 19, posted (11 years 9 months 1 week 2 days 7 hours ago) and read 834 times:

NEGATIVE!

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