My answers seem ok but differ from book.

[jiboom]

1) a car moves along a straight road against a resistance to motion which is 10 times the speed of the car. the car has a mss of 800kg and the engine is working at the constant rate 5kw. find an expression for the acceleration of the car.

my attempt:

P=driving force.v => 5000/v=driving force

f=ma=>800a=5000/v-10v =>a=50/8v -v/80

book answer is 5/v-v/80



2) a particle leaves a point A at time t=0 with speed u and moves towards a point B with retardation Lv,where v is the speed of the particle at time t. the particle is at a distance s from A at time t.

i have then shown:

v=u-Ls...(1)

ln(u-Ls)=ln u-Lt...(2)



at t=o a second particle starts from rest at B and move towards A eith accel 2+6t. the particles collide at the midpoint of AB when t=1. find the distance AB and the speeds of the particles on impact.

here i found AB=4 and speed of particle from B is 5 to the left.


for speed of A i have used (1) (2) with t=1 s=(AB/2)=2

(2) gives ln(u-2L)=lnu-L

L=ln(u/u-2L)


e^L=u/u-2L

u=-2L/(1-e^L)

then (1) gives

v=u-2L
=-2L[{1/(1-e^l)}+1]

=-2L/(1-e^L)[1+1-e^L]

=2L{e^L-2}/(1-e^L}

but book gives

-2L(1-e^L}


3)

a disc is set spinning and turns through T degrees in t seconds where

T=120t-6t^2,

calculate the rate in revolutions per minute,at which the disc is rotating when t=6

im saying (guesing)

T=(120t-6t^2)∏/180 gives me T in radians

DT/dt=ω=(120-12t))∏/180

when t=6,

ω=48∏/180 rads/6 secs

so revs per min =

2∏*10/(48∏/180)

=1800/24=75

but books says 8.

im sure I am going wrong on this one!

 

[BruceW]

I haven't fully looked through your answers, but there's a couple of things to watch out for: Firstly, the equations are not dimensionally correct. Which means that some assumption must have been made about which system of units must be used. So if you are using a different system of units than the book, then this might cause you to get a different answer to them. Secondly, the phrases "resistance to motion" and "retardation" are not immediately obvious what they mean physically. So maybe you have taken them to mean something different to what the book means. Hope this advice has helped.

 

[jiboom]

I haven't fully looked through your answers, but there's a couple of things to watch out for: Firstly, the equations are not dimensionally correct. Which means that some assumption must have been made about which system of units must be used. So if you are using a different system of units than the book, then this might cause you to get a different answer to them. Secondly, the phrases "resistance to motion" and "retardation" are not immediately obvious what they mean physically. So maybe you have taken them to mean something different to what the book means. Hope this advice has helped.

which one (s) do you refer to here?

im very confident (1) is incorrect. the very next question is the same but has no numbers. it has mass m,constant rate homework and resistance to motion kv then asks to show acc = blah blah. this i can do and when i put the nunmbers from my question in i get my answer not the books



any takers to (2) and especially (3) ?

for (2) the book answer is what u ,the initial velocity,equals using equation (2) so can't believe that is velocity when they collide so I am pretty sure the book is wrong here too.

 

[BruceW]

I've had a proper look through question 3), and it says that time must be in seconds, and you have done that, so there should be no problem there. I think you went wrong when you got to here:

ω=48∏/180 rads/6 secs

Its almost right, but it should be 'per secs', not 'per 6 secs'. Since the angular speed is the instantaneous change, the 6 should not be there. So this makes the rest of the calculation incorrect. Also, in the next step, you multiply by 2pi to get the revolutions. But it should be the other way around, since angular speed = 2pi times frequency.

 

[jiboom]

from
ω=48∏/180 rads/6 secs

i was being thick there i guess:

T was quoted as degrees per t seconds so i just put t=6 to get /6 secs.

for second part i did mean to do ω/2∏ doh!

so ignoring the 6 i would have

ω/2∏= (48∏/180) /2∏

=24/180 per second so need

60*24/180

= 8

thaks for that,silly mistake with that one. would appreciate it if you could look over 1,2 to see if there is similar errors (saying that,as mentioned above (1) looks wrong from a following question. it seems they wanted power to be 4kw)

 

[BruceW]

OK, looking at question 2), I'm not sure what the retardation means, and I'm not sure how you got those two equations, but assuming that they are correct:

I'd say the distance AB=4 is correct. And you're right that the speed of the second particle at the halfway point is 5 to the left.

I think you went wrong on this step:

e^L=u/u-2L

u=-2L/(1-e^L)

I think maybe you didn't rearrange it right?

 

[jiboom]

OK, looking at question 2), I'm not sure what the retardation means, and I'm not sure how you got those two equations, but assuming that they are correct:

I'd say the distance AB=4 is correct. And you're right that the speed of the second particle at the halfway point is 5 to the left.

I think you went wrong on this step:

I think maybe you didn't rearrange it right?

e^L=u/u-2L

yeah, i should have

u=-2Le^l/(1-e^L)

v=u-2L
=-2L[{e^l/(1-e^l)}+1]

=-2L/(1-e^L)[e^l+1-e^L]

-2L/(1-e^L}

thanks. can't believe i did not see that! just a case of looking to hard and becoming blind

 

[BruceW]

yeah, that's happened to me many times as well. It kind of reminds me how in chess, you're looking at the position with a certain preconception, and you can look at it for ages without seeing the obvious correct move. Which is why, in chess, you need to take a step back and look at the board differently, every so often.

So do you get the right answer using that? And you were saying that you got question 1) wrong because you got an earlier part wrong?

 

[jiboom]

dont get me started about chess ha ha. i have a knack of only seeing why a move is bad after i have moved. every game i see what looks a good move,make it then almost instantly see the refutation!

any way,question 1) the book is wrong I am sure. i would appreciate it if you agreed with me though. as i say,the next question in the book was identical except it didnt have numerical values. when i take that answer and put the numbers in for 1) i get what i suggest not the book answer.

 

[BruceW]

For question 1), I get the same answer as you. The question isn't completely clear, so I assumed that: 'resistance to motion' is a force, and that all the units were SI. As long as these two things are true, I think you got the right answer.

Yeah, I do that all the time in chess. I guess the first step to getting better is to realize what your bad habits are!