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

"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:**

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.

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

"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!

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)

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)

Thanks Fritzi

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

JSC_23

-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!

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

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

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

Its the same formula.

Its the same formula.

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

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

Thanks for you help...

JSC_23

Be a realist...Remain a dreamer!

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.

Sorry if I mislead you. As I said, I am under a lot of pressure right now.

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

v=u + 2as

v=0ms

u=40ms

s=100m

0=40 + 2(a)(100)

-40 = 200a

a=-5ms^-2

Therefore acceleration =

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

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

JSC_23

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

JSC_23

Be a realist...Remain a dreamer!

BEST OF LUCK

for your exam Fritzi...i hope everything goes well!!!!!!!!!

JSC_23

for your exam Fritzi...i hope everything goes well!!!!!!!!!

JSC_23

Be a realist...Remain a dreamer!

- delta-flyer
**Posts:**2631**Joined:**

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

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"

"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!

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:**

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"

- delta-flyer
**Posts:**2631**Joined:**

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

..... speechless?!!

..... speechless?!!

"In God we trust, everyone else bring data"

NEGATIVE!

Remember, if you are__decelerating__ your answer must be a __negative__ number!

redngold

Remember, if you are

redngold

Up, up and away!

Users browsing this forum: No registered users and 21 guests