Question

A particle moves along the x-axis with the velocity given by v(t)=3t/(1+t^2) for t >or equal to 0. When t=0, the particle is at the point (4,0). 1. Determine the maximum velocity for the particle. Justify your answer. 2. Determine the position of the particle at any time t. 3. Find the limit of the velocity as t->infinity. 4. Find the limit of the position as t->infinity.

Answer 1

dx = vdt

use this..this little thing is supposed to do wonders for you ! :D

Answer 2

Could you explain how that formula works please?

Answer 3

have you studied integration ?

Answer 4

Just started learning it. It's not my best subject. xD

Answer 5

the definition of velocity is
v=dx/dt

Answer 6

1.
a(t) = 3(1-t^2)/(1+t^2)^2
max v at a=0, or t=1
v(1) = 3/2

2.
x(t) = 3/2 log(1+t^2)+C
at t=0, x=4, so
4 = 3/2 log(1+0) + C
C = 4
x(t) = 3/2 log(1+t^2) + 4

3.
v(t) → 0 as t→∞

4.
x(t) → ∞

that's what I've got so far. Does that make sense?

Answer 7

@sauravshakya

Answer 8

you sure differentiated right in the 1st part..
rest all seem perfect! :)

Answer 9

At least someone agrees with it so far.

Answer 10

i missed a question mark there :P
i meant
you sure differentiated right in the 1st part ????
:D

cause i see that wrong..please confirm..

Answer 11

ohh wait//thats -t^2 //
sorry sorry..thats correct..

Answer 12

my apologies! :P

Answer 13

Yeah I just checked. oh well haha

Answer 14

Could somebody please check this to make sure it makes sense?

1. dv/dt = [(1+t^2)(3) - 3t(2t)] / (t^2+1)^2
=[3 t^2 + 3 - 6 t^2 ] /(t^2+1)^2
zero when t= 1 (since t always >/=0)
so the max v is at t = 1
then at t = 1
v(1) = 3/2

2.
x(t) = 3/2 log(1+t^2)+C
at t=0, x=4, so
4 = 3/2 log(1+0) + C
C = 4
x(t) = 3/2 log(1+t^2) + 4

3. as t --> oo, v--> 3t/t^2 = 3/t = 0

4. as t -->oo, x --> (3/2)t^2 = oo