Velocity with respect to displacement (1 Viewer)

[alez]

Not sure on how to change the velocity (in terms of x) to terms of t.

A particle moving in a straight line has a constant acceleration of -10m/s^2. It is initially projected from the origin with a velocity of 12m/s.
i-Find v^2 as a function of x
ii-Find where the particle comes to rest
iii-What is the velocity when the particle returns to the origin
iv-Show that v can be expressed as a function of t by v=12-10t
v-At what time does the particle return to the origin

Having trouble with part iv. Not sure how to go about it
I found v with respect to x, v^2=-20x+144

 

[PC]

Try integrating with respect to t.

We know that a = –10
Since initial velocity is 12 m/s, we have v = –10t + 12, or v = 12 – 10t.
Since the particle starts at the origin, we have x = –5t² + 12t.

Don't forget the simple stuff!

 

[alez]

But that doesnt show how v can be expressed as a function of t

 

[independantz]

But that doesnt show how v can be expressed as a function of t

What do you mean he got the exact same answer as required in the question, i.e v=12-10t and this shows that v is being expressed as a function of t.

 

[vds700]

But that doesnt show how v can be expressed as a function of t

As the acceleration given is a constant, it is perfectly OK to integrate with respect to x or t, depending on what u nned. You are not expected to do
dx/dt = sqrt(144 - 20x)
dt/dx = 1/sqrt(144 - 20x)

then integrate with respect tio x to find t as a function of x and rearrange etc. It is far easier to just integrate -10 with respect to t to find the veloacity.

 

[conics2008]

Not sure on how to change the velocity (in terms of x) to terms of t.

A particle moving in a straight line has a constant acceleration of -10m/s^2. It is initially projected from the origin with a velocity of 12m/s.
i-Find v^2 as a function of x
ii-Find where the particle comes to rest
iii-What is the velocity when the particle returns to the origin
iv-Show that v can be expressed as a function of t by v=12-10t
v-At what time does the particle return to the origin

Having trouble with part iv. Not sure how to go about it
I found v with respect to x, v^2=-20x+144

use this if x ( double dot ) is given in terms of x ..
then Velocity = 1/2 v^2 = S f(x)

i) x=0 x( dot) = 12 x(doubel dot ) = -10

1/2 v^2= S -10 >> 1/2v^2=-10x + C

now when x= 0 x(dot) = 12 >> C=12

x(dot) = 1/2v^2=-10x+12 >>>>> v^2=-100x+24

ii) v=0 >> 100x=24 >>> when x= 24/100

iii) make x=0 >> v^2=24 v=root of 24

iiii) v=root of 24-100x

dx/dt=root of 24-100x >> dt/dx= 1/ root of 24-100x

therefore t= S 1/ root of 24-100x

you integrate this. i cant be stuffed.

when you do this find your C by subbin in x=0 t=0

v) sub in x=0 and find t...

good day i understood this, hope you can to.

PS: I did not simplify any of my expression, because i just did it right now on top of my head without any paper or calculator.. you should simplify every step to minimise confusion and errors.

 

[alez]

Ok thanks. The wording of the question was confusing me. I managed to make it excessively complicated by integrating dt/dx and having to use substitution....
It was quite messy
But thanks I get it now