here's the question:

Q1. An driver in an XFR exits the first turn in an XFR on Blackwood GP at 80k/h. He then accelerated at 8m/s for 10 seconds, how far has he travelled during that period?

Q2. A Formula XR starts from rest with an acceleration of 10m/s , find (a) the distance travelled in 10s and (b) the deacceleration needed to bring the Formula XR to rest

I was looking back at some of my high school books today, and the physics textbook was quite interesting, so i decided to post two questions up on here, can you solve them?


cheers

m/s is a velocity, not an acceleration.
And 80.000 1/h is not a velocity.

(The only way to learn to use proper units is to be laughed at. )

Vain

A1. Assuming you mean 80 km/hr (~22.22 m/s) and 8 m/s^2 for velocity and acceleration, respectively:

distance travelled = vt + .5at^2 = (22.22*10) + (.5*8*10*10) = 622.2 m

A2. Assuming 10m/s^2

(a)distance travelled = vt + .5at^2 = (0*10) + (.5*10*10*10) = 500 m
(b)can have several answers. You didn't state any time limit, so really the car could take 20 minutes to come to rest if the driver wants to, or a matter of seconds. You'd need to ask for a magnitude of "deacceleration" (which is still actually an acceleration ;-) ) to come to rest within so many seconds.

I think he means this:

Q1. A driver in an XFR exits the first corner on Blackwood GP at 80kmh^-1 (22.22 ms^-1). He then accelerates at 8ms^-2 for 10 seconds, how far has he travelled during that period?

Q2. A Formula XR starts from rest with an acceleration of 10ms^-2 , find (a) the distance travelled in 10 seconds and (b) the deceleration needed to bring the Formula XR to rest from the speed it gained after 10 seconds.

Edit: Miles too late

Quote from efast :

can you solve them?

yes

I bet having a brain is ace.

suvat equations don't you just love them lol.

Quote from efast :

Q2. A Formula XR starts from rest with an acceleration of 10m/s , find (a) the distance travelled in 10s and (b) the deacceleration needed to bring the Formula XR to rest

...

can you solve them?

I can solve the answers much more accurately than that.

Quote from efast :

here's the question:

Q1. An driver in an XFR exits the first turn in an XFR on Blackwood GP at 80k/h. He then accelerated at 8m/s for 10 seconds, how far has he travelled during that period?

Q2. A Formula XR starts from rest with an acceleration of 10m/s , find (a) the distance travelled in 10s and (b) the deacceleration needed to bring the Formula XR to rest

I was looking back at some of my high school books today, and the physics textbook was quite interesting, so i decided to post two questions up on here, can you solve them?


cheers

Q1. Answer: not very far. the car lost grip and he ate it in the chicane

Q2. you know someone else asked me if I knew much about the Laws of Physics...I never went to law school or been to Physics, so I just don't know

The driver of an LX4 presses the throttle until 9000rpm is reached and quickly releases it. A particular point's y-coordinate on the exhaust vibrates in such a way that it satisfies the differential equation:

(d^2y/dt^2) - (dy/dt) - 2y = 10 sin t

Where t is in ms after the release of the throttle and y is mm relative to the rest position of the exhaust. Determine the solution for y which remains finite as t -> infinity given that x = 4 when t = 0.

My cat's name is Mittens.

here's the answer:

Q1. Do your own homework.

Q2. See above.

Quote from efast :

here's the question:

can you solve them?

Go to the end of the book no doubt and look at the answers

Quote from axus :

The driver of an LX4 presses the throttle until 9000rpm is reached and quickly releases it. A particular point's y-coordinate on the exhaust vibrates in such a way that it satisfies the differential equation:

(d^2y/dt^2) - (dy/dt) - 2y = 10 sin t

Where t is in ms after the release of the throttle and y is mm relative to the rest position of the exhaust. Determine the solution for y which remains finite as t -> infinity given that x = 4 when t = 0.

y = -3*sin(t) + cos(t) + k1*e^(2*t) + k2*e^(-t)

with k1 and k2 the two integration constants. (No, I'm not that good at solving differential equations: I just typed it in into Maxima ). Now, I might be wrong but it looks to me like the only way to keep y finite is to lose the term with e^(2*t). Therefore k1 = 0.
Now I'll assume you meant 'y = 4 when t = 0' instead of 'x = 4 when t = 0' (since x is entirely irrelevant anyway). Substituting k1 = 0, y = 4 and t = 0:

4 = -3*sin(0) + cos(0) + k2*e^0

or

k2 = 4 + 3*sin(0) - cos(0) = 3

So:

y = -3*sin(t) + cos(t) + 3*e^(-t)

Quote from Greboth :

suvat equations don't you just love them lol.

yup, oh well, i'm now revising for my GCSE's!

Atleast its not like Jakg reciting pi in TS...

Quote from vrt3 :

y = -3*sin(t) + cos(t) + k1*e^(2*t) + k2*e^(-t)

with k1 and k2 the two integration constants. (No, I'm not that good at solving differential equations: I just typed it in into Maxima ). Now, I might be wrong but it looks to me like the only way to keep y finite is to lose the term with e^(2*t). Therefore k1 = 0.
Now I'll assume you meant 'y = 4 when t = 0' instead of 'x = 4 when t = 0' (since x is entirely irrelevant anyway). Substituting k1 = 0, y = 4 and t = 0:

4 = -3*sin(0) + cos(0) + k2*e^0

or

k2 = 4 + 3*sin(0) - cos(0) = 3

So:

y = -3*sin(t) + cos(t) + 3*e^(-t)

You can dress it up as LFS all you like, but it's still a maths textbook question - and I gave them up 15* years ago... Now I just do heuristics, long winded algebra, and conditional statements - but the difference is I do so voluntarily - not because some bald headed guy wants me to!

*As this is a maths thread I should point out that it is 15 +/- 2.

Quote from Becky Rose :

You can dress it up as LFS all you like, but it's still a maths textbook question - and I gave them up 15* years ago... Now I just do heuristics, long winded algebra, and conditional statements - but the difference is I do so voluntarily - not because some bald headed guy wants me to!

A couple of years back I found myself in the embarrassing situation of having to look up some GCSE-level maths stuff on the web in order to do a job I'd got in, and it was endlessly amusing to me that I actually had needed to know that stuff in real life, so I emailed my old school to tell them that they were right all along.

Are you kidding me? Is this supposed to be a challenge?

Quote from axus :

(d^2y/dt^2) - (dy/dt) - 2y = 10 sin t

Go away!
I've spent all day solving DEs because I write an exam on them next monday, so just go away with it, will ya? I'm trying to avoid maths here right now!


Vain

I have a PhD in Mathematics

Hi btw, 2nd post from me

Quote from Vain :

m/s is a velocity, not an acceleration.
And 80.000 1/h is not a velocity.

(The only way to learn to use proper units is to be laughed at. )

Vain

wow....i will get laughed at by physics teachers if i put down what you wrote

Quote from NetDemon01 :

A1. Assuming you mean 80 km/hr (~22.22 m/s) and 8 m/s^2 for velocity and acceleration, respectively:

distance travelled = vt + .5at^2 = (22.22*10) + (.5*8*10*10) = 622.2 m

A2. Assuming 10m/s^2

(a)distance travelled = vt + .5at^2 = (0*10) + (.5*10*10*10) = 500 m
(b)can have several answers. You didn't state any time limit, so really the car could take 20 minutes to come to rest if the driver wants to, or a matter of seconds. You'd need to ask for a magnitude of "deacceleration" (which is still actually an acceleration ;-) ) to come to rest within so many seconds.

good job, netdemon, nice workings

one person got it rite at least, others either forgot about their high school studies or have no idea,

Quote from DaveWS :

I think he means this:

(b) the deceleration needed to bring the Formula XR to rest from the speed it gained after 10 seconds.

i did forget to put in the time to find deacceleration to bring the formula XR to rest, good to see you corrected it


and thanks to everyone who "tried" to answer the question

Seeing these equations literally elicits the fight or flight response in me and makes me want to run away from my computer while throwing things at it to make sure it doesnt chase me.

Quote from Lautsprecher[NOR] :

My cat's name is Mittens.

My cat's breath smells like cat food.