Question

A particle moving with a constant acceleration describes in the last second of it motion 9/25th of the whole dist.
if it starts from rest,how long is the particle is the particle in motion and through what dist does it movie if it describes 6cm in the first sec?

Answer 1

is this calculus ?

Answer 2

Kinematics.

Answer 3

And yes,I did learn some Calculus.

Answer 4

do you have a formula for distance as a function of acceleration ?

Answer 5

S = ut + 0.5at^2

Answer 6

ok, and they say
it starts from rest
that tells us what u is. any idea ?

Answer 7

If you the body is at rest ,then u = 0 ms^-1.

Answer 8

Therefore,S = 0.5at^2

Answer 9

sounds good.

it describes 6cm in the first sec
if we use that info, what do we get for a ?

Answer 10

Idk.
Thats where I'm stuck.
I have no idea to move on any further.

Answer 11

you know this
S = 0.5at^2

you know
it describes 6cm in the first sec
which you have to interpret: it moved how far? how long did it take ?
that info corresponds to your equation: s and t.

Answer 12

I'm so confused as idk what to do.. e.e...

Answer 13

do i need to take the distance travaled as 9/25 in t-1 seconds?

Answer 14

"describe" is not the clearest word (the 5th or 6th definition is "trace the outline"
I would use the word "traveled"

Answer 15

right now, we are concentrating on these two things:

s = 0.5 a t^2
and
it traveled 6 cm in the first second.

Answer 16

so then
0.06 = 0.5*a*1
a = 0.012ms^-2?

Answer 17

yes
though maybe it's easier to work in cm rather than meters to avoid the decimals
we can switch to meters at the end if we have to.

Answer 18

now we know what a is
s= 0.5 * a * t^2
s= 6 t^2 (in cm , not meters)

Answer 19

it should be 6*10^-2 m?

Answer 20

yes, it can be written in meters. In that case,
s= 0.06 t^2 (in meters)

Answer 21

why didnt you include a in the equation?

Answer 22

we started with
s = 0.5 a t^2
we put in 0.06 m for s, and t=1:
0.06 m = 0.5 *a *1 sec^2
mulitply both sides by 2
0.12 m/sec^2 = a

Answer 23

now that we know a, we can write
s= 0.5 * a * t^2
and put in a;
s= 0.5*0.12*t^2
and simplify that to
s = 0.06 t^2
that is the formula for how far the particle moves

as a check: after 1 second we get
s= 0.06 * 1 = 0.06 m= 6 cm
which is what it should be.
ok ?

Answer 24

okay.

Answer 25

What next?

Answer 26

we want to find
how long is the particle is the particle in motion

which means find "t" (how long it moved)

we have to use the last info they gave us.
describes in the last second of it motion 9/25th of the whole dist.

Answer 27

so

when t-1

then
s = 0.5a(t-1)^2?

Answer 28

I'm wondering what is the nicest way to to think about it.
we know a few things
if we let the total amount of time be T
then
s(T) - s(T-1) = (9/25) s(T)

s(T) is short for "how far it moved in T seconds"

Answer 29

s(t-1) = 9/25 of S

Answer 30

no.
it moves in the last second of it motion 9/25th of the whole distance
the "whole distance" is s(T)
if it moved 9/25 of the distance in the last second, that means it moved 16/25 of the distance in the time from 0 up to T-1

in other words, the distance it moves in time T-1 is 16/25 of the entire distance
s(T-1) = 16/25 s(T)

Answer 31

s(T-1) = 16/25 s(T)
or
25 s(T-1) = 16 s(T)

Answer 32

howd you get 15?typo?

Answer 33

somehow I changed 25 into 15. gack!

Answer 34

can you solve the "quadratic equation" that we get ?

Answer 35

25 s(T-1) = 16 s(T) is not an quadratic eqn....

Answer 36

25(t^2 + 2t+1) = 16

Answer 37

remember
s(T) is short for 0.06*T^2
and
s(T-1) is short for 0.06*(T-1)^2

Answer 38

25*0.06*(t-1)^2 = 16*0.06*t^2

so 0.06 get cancelled?

Answer 39

yes the 0.06 cancels

we have to expand (t-1)^2
collect terms

Answer 40

t^2 -2t + 1

Answer 41

ok, put that into the equation to replace (t-1)^2

Answer 42

25t^2 -50t + 25

Answer 43

and don't forget the other side

Answer 44

9t^2 - 50t + 25 = 0

Answer 45

I used the quadratic formula, but it does factor.

Answer 46

i got 5/9 and 5

Answer 47

that is the time in seconds.
we know it goes at least 1 second, so the first solution 5/9 is not relevant

Answer 48

t = 5s

Answer 49

and now you find the total distance s = 0.06 t^2

Answer 50

1.5 m

Answer 51

oops...1.5 cm

Answer 52

no , it's meters

remember we are using 0.06 m/sec^2

Answer 53

so is the final ans 1.5 m?

Answer 54

and also
it describes 6cm in the first sec?
so you know the answer can't be 1.5 cm (which is smaller than 6 cm)

yes, the answer is
it takes 5 seconds, and travels a total distance of 1.5 meters

Answer 55

ty