Question

Figure 2-25

depicts the motion of a particle moving along an x axis with a constant acceleration. The figure's vertical scaling is set by xs = 7.20 m. What are the (a) magnitude and (b) direction of the particle's acceleration ( +x denote 1 , 0 - otherwise )?

Answer 1

@.Sam. @amistre64 could someone please help. I dont know how to figure this out.

Answer 2

we are moving along the x axis correct?

Answer 3

we have 3 points to define the position with such that

p(t) = 1/2 at^2 +vo t +po

Answer 4

yes

Answer 5

po is the initial position = -1

but it might be more useful in just makeing the 3 obvious points fit

Answer 6

1/2 a 0^2 +vo 0 +po = -1
1/2 a 1^2 +vo 1 +po = 0
1/2 a 2^2 +vo 2 +po = xs

Answer 7

po = -1

1/2 a +vo = 1
2a +2vo = xs + 1

-a -2vo = -2
2a +2vo = xs + 1
-----------------
a = xs - 1

Answer 8

But the initial position is not -1 since we don't know how the position axis is scaled

Answer 9

hmm, your right i thought i saw it but that was my own dementia :)

Answer 10

its alright, I thought it was -1 too :)

Answer 11

is it fair to say its xs/3 ?

Answer 12

or can we trust the grid?

Answer 13

if it was xs/3...would that mean constant velocity?

Answer 14

let me do a more careful reading of this ...

graph depicts the motion of a particle moving along an x axis with a constant acceleration.

if a = k
v = kt + c
p = 1/2 kt^2 +ct + n


The figure's vertical scaling is set by xs = 7.20 m.
and at t=0 we appear to be 1/3 of xs


What are the
(a) magnitude and
(b) direction of the particle's acceleration ( +x denote 1 , 0 - otherwise )?

Answer 15

the points of interest that i see are: (0,xs/3); (1,0); (2,xs)

Answer 16

ok

Answer 17

1/2 k 0^2 +c0 + n = -xs/3; n = -xs/3

1/2 k 1^2 +c1 = 0 + xs/3

1/2 k 2^2 +2c = xs + xs/3

2k +2c = xs + xs/3

------------------------
1/2 k + c = xs/3
2 k +2c = xs + xs/3

now we can solve for k, and c by a system of equations

- k -2c = -2xs/3
2 k +2c = xs + xs/3

k = xs - xs/3 = 4.8
.................................

2(4.8) +2c = 9.6

4.8 +c = 4.8

c = 0

Answer 18

Sorry, I use a bit different notation...I think k is constant, but what is c and n?

Answer 19

p = 2.4 t^2 - 2.4

Answer 20

well, since i used a k for a constant, i had to think of other letters to define different constants

k=a, c=vo, and n is i forget the physics position constant .. so maybe?

Answer 21

oh,ok

Answer 22

so how dies this information proved solutions?

Answer 23

What are the
(a) magnitude and 4.8
(b) direction of the particle's acceleration: in the +x direction whish is (1,0) ??

( +x denote 1 , 0 - otherwise )
^^^^^ whats all this stuff mean?

Answer 24

that just means to denote positive acceleration with a 0 and everything else is just denoted with a 0.

Answer 25

if we had to define a vector id say it was: accel. = 4.8(1,0)

Answer 26

Ok, I am still working through the info you gave me up there and converting it to make sense for me.

Answer 27

good luck with that, im still doing the same :)