Question

AP maths

a particle moves along the positive x-axis, at time t seconds after leaving point O from rest, the displacement of the particle from O is x cm. The acceleration is defined by

a= {5-3t , 0 < or equal t < or equal 1
-(4x+1), t>1

find

(a) the max velocity of the particle and its distance from O when this happens,

(b)the speed of the particle when x=5/2.

Answer 1

a=v(dv/dx) or dv/dt 0r d^(2)x/dt^(2)

Answer 2

use a=dv/dt for first second and a=v(dv/dx) from later on
to find expressions foe velocities
now find the maximums in each case. the maximum of the wo gives the maximu velocity

Answer 3

You should get 1/2 ms^-1 , I think.
At x = 1/4m

Answer 4

ya mistake so i delet the post..

Answer 5

Yeah. Got it. Didnt refresh the page :P

Answer 6

for d second equation right??

Answer 7

a=0
-(4x+1)=0
4x+1=0
x=-1/4

Answer 8

\[a=5-3t\]
\[v=5t-\frac{3}{2}t^2+c\] i guess can we find c?

Answer 9

c=0 bcoz when x=0, t=0, v=0

Answer 10

oh starts at rest so i suppose c = 0

Answer 11

and for \(t> 1\) we have \(a=-(4t+1)=-4t-1\) which confuses me a bit because it seems we have a jump discontinuity in acceleration, a phenomenon rare in nature (nature being by in large continuous at the observable level)

Answer 12

ya.. what u say is correct.. bcoz my sch teacher always comment that the local book are more on theory only instead of giving real situation

Answer 13

ok no matter we have
\[v(t)=-2t^2-t+c\] and as for c maybe we should at least make sure that the velocity in continuous

Answer 14

for the second equation.. t>1 when a=-(4x-1) not t.... thn wwe need to use a=v(dv/dt) to solve the equation...

Answer 15

we*

Answer 16

wait acceleration is a function of x?

Answer 17

ya for t>1

Answer 18

but for t is in between [0,1] we use acceleration as a function of t

Answer 19

then you need so solve a second order differential equation
what class is this?

Answer 20

you have \(x''(t)=-4x(t)+1\)

Answer 21

differential equation the application part

Answer 22

ooooooooooooooooooooooooooooh then i guess we have to solve that one yikes
i am sure it is easy and i am sure i don't know how to do it right off the bat
any ideas?

Answer 23

I dont think we have to integrate twice
use a=vdv/dx

Answer 24

ok go nuts. i think it will be some combination of sine and cosine, but since you have the velocity, which is the second derivative, as a function of position, i think you have to sole a second order equation
i could easily be wrong

Answer 25

\[v ^{2}=2 \int\limits_{i}^{f}a dx\]

Answer 26

i think i can solve the part (a) but it is very long.....

Answer 27

max velocity is dv/dt=0
==> a=0

for t is in between [0,1]
0=5-3t
t=5/3

but max t=1



let
when t=1, v=v1
when t=0,v=v0
therefore v1>v0

therefore for the equation to valid, v1=max v

a=5-3t
dv/dt=5-3t
integrate dv=intergrate 5-3t dt
v=5t-3t^(2)/2+c
when t=0,x=0,v=0
so c=0
v=5t-3t^(2)/2

at t=1
v = v max = 5(1)-3(1)^(2)/2 =7/2