https://imgur.com/a/uALAK16

what am i doing wrong

average the speeds and times it by the time (I have no idea what this guys homework is saying)

melon_fan [#2]

average the speeds and times it by the time (I have no idea what this guys homework is saying)

oh wait speed falls off right?

Ideally you divide the 20s into 3 parts:

1. 0-10s (a = 2m/s^2)
2. 10-15s (a = 0m/s^2)
3. 15-20s (a = -3m/s^2)

you just figure out the velocity of each one and then use the time to get the distance and then add all 3 up

hit me up if you need further help

CryoZanderDerrekEnjoyer [#4]

Ideally you divide the 20s into 3 parts:

1. 0-10s (a = 2m/s^2)
2. 10-15s (a = 0m/s^2)
3. 15-20s (a = -3m/s^2)

you just figure out the velocity of each one and then use the time to get the distance and then add all 3 up

hit me up if you need further help

that’s basically what I was saying

EpicGamerGamingEpicly [#5]

222.5

this is wrong

answer is Botssi

CryoZanderDerrekEnjoyer [#4]

Ideally you divide the 20s into 3 parts:

1. 0-10s (a = 2m/s^2)
2. 10-15s (a = 0m/s^2)
3. 15-20s (a = -3m/s^2)

you just figure out the velocity of each one and then use the time to get the distance and then add all 3 up

hit me up if you need further help

DISTANCE: s = 1/2a x t^2 + v0 x t

VELOCITY: s = v^2 - v0^2 / 2a

s stands for distance

part1: 0 - 10s

s: 1 x 100 = 100m
100m = s

v: 100 = v^2 / 4
v: 400 = v^2
20m/s = v

part2: 10 - 15s

s: 0 x 25 + 20 x 5
s: 20 x 5
100m = s

xxvelocity stays same as no acceleration (v = 20m/s)xx

part3: 15 - 20s

s: -1.5 x 25 + 20 x 5
s: -37.5 + 100
62.5m = s

part1 + part2 + part3 = 100m + 100m + 62.5m + = 262.5m

CryoZanderDerrekEnjoyer [#9]

DISTANCE: s = 1/2a x t^2 + v0 x t

VELOCITY: s = v^2 - v0^2 / 2a

s stands for distance

part1: 0 - 10s

s: 1 x 100 = 100m
100m = s

v: 100 = v^2 / 4
v: 400 = v^2
20m/s = v

part2: 10 - 15s

s: 0 x 25 + 20 x 5
s: 20 x 5
100m = s

xxvelocity stays same as no acceleration (v = 20m/s)xx

part3: 15 - 20s

s: -1.5 x 25 + 20 x 5
s: -37.5 + 100
62.5m = s

part1 + part2 + part3 = 100m + 100m + 62.5m + = 262.5m

I actually looked at it it should be right (I’m a freshman idk what I’m doing)

It's in 250-300

fuck physics why does it exist im gonna fail the AP test

CryoZanderDerrekEnjoyer [#9]

DISTANCE: s = 1/2a x t^2 + v0 x t

VELOCITY: s = v^2 - v0^2 / 2a

s stands for distance

part1: 0 - 10s

s: 1 x 100 = 100m
100m = s

v: 100 = v^2 / 4
v: 400 = v^2
20m/s = v

part2: 10 - 15s

s: 0 x 25 + 20 x 5
s: 20 x 5
100m = s

xxvelocity stays same as no acceleration (v = 20m/s)xx

part3: 15 - 20s

s: -1.5 x 25 + 20 x 5
s: -37.5 + 100
62.5m = s

part1 + part2 + part3 = 100m + 100m + 62.5m + = 262.5m

im somewhat getting it

are the equations you're using given kinematic equations?

melon_fan [#10]

I actually looked at it it should be right (I’m a freshman idk what I’m doing)

idk man maybe i did a error here or there but this should be good (physics is my best subject)

RickyIndian [#13]

im somewhat getting it

are the equations you're using given kinematic equations?

yessir

1phnxe [#12]

fuck physics why does it exist im gonna fail the AP test

If its not too difficult physics then hit me up, i can look what i can do.

more help pls

there was a day I could solve this in 1 minute

bump

not webassign homework LMAO

FireII [#20]

not webassign homework LMAO

how did bro know LOL

RickyIndian [#21]

how did bro know LOL

I'm 15 bro i have calculus to do as well 💀

-6.344 cm/s2

Ask chatgpt 🗿

−11.6 cm/s^2 is what i got but idk if i did the math correct. the idea should be right though.

i used the one of the uniform acceleration equations, displacement = velocity initial time + (1/2) acceleration * time^2 (d = vit + (1/2)at^2). Everything is given to you- the displacement in the x-direction is xf-xi. you have the time interval and initial velocity, so just solve for acceleration.

It's pretty simple with the four main kinematic constant acceleration equations, you just have to get the idea down.
iirc vi*t calculates the intial displacement at the initial velocity over the time interval. But because the object is accelerating, you are gaining more displacement at the increasing speeds over the time interval. Think of it as adding the original displacement of the object if it were traveling at the initial speed (11 cm/s) for 2.45s. But since the object gains velocity from that initial velocity at an acceleration, 1/2at^2 calculates such displacement.

imagine the object traveled at 11cm/s for 2.45s and displaced 26.95 cm. Now imagine we travel at the same speed and displacement right next to the object, treating speed relative. We are both currently traveling at a constant velocity. the object in this scenario gains an imaginary little hand that applies a force and accelerates it. We would begin to see the acceleration because the object gains velocity (in this example we would see the object slow down relative to us because it accelerates in the negative x-direction). That negative displacement from the acceleration is the 1/2at^2 we add to the original displacement at 11cm/s for 2.45s.

CryoZanderDerrekEnjoyer [#4]

Ideally you divide the 20s into 3 parts:

1. 0-10s (a = 2m/s^2)
2. 10-15s (a = 0m/s^2)
3. 15-20s (a = -3m/s^2)

you just figure out the velocity of each one and then use the time to get the distance and then add all 3 up

hit me up if you need further help

did he edit the linked image what you are doing looks nothing like the question there?

DevoidDecoy [#25]

−11.6 cm/s^2 is what i got but idk if i did the math correct. the idea should be right though.

i used the one of the uniform acceleration equations, displacement = velocity initial time + (1/2) acceleration * time^2 (d = vit + (1/2)at^2). Everything is given to you- the displacement in the x-direction is xf-xi. you have the time interval and initial velocity, so just solve for acceleration.

It's pretty simple with the four main kinematic constant acceleration equations, you just have to get the idea down.
iirc vi*t calculates the intial displacement at the initial velocity over the time interval. But because the object is accelerating, you are gaining more displacement at the increasing speeds over the time interval. Think of it as adding the original displacement of the object if it were traveling at the initial speed (11 cm/s) for 2.45s. But since the object gains velocity from that initial velocity at an acceleration, 1/2at^2 calculates such displacement.

imagine the object traveled at 11cm/s for 2.45s and displaced 26.95 cm. Now imagine we travel at the same speed and displacement right next to the object, treating speed relative. We are both currently traveling at a constant velocity. the object in this scenario gains an imaginary little hand that applies a force and accelerates it. We would begin to see the acceleration because the object gains velocity (in this example we would see the object slow down relative to us because it accelerates in the negative x-direction). That negative displacement from the acceleration is the 1/2at^2 we add to the original displacement at 11cm/s for 2.45s.

You're correct I just think you didn't step it out in a very clear way.
You should start most problems like this by writing out the information you have in a clear way so you can figure out what formula you want to use.
In this case we have: t = 0s, where vi = 11cm/s, and d = 2.91cm
then we also have : t = 2.45s , vf = ? and d = -5 cm (note we don't need the final velocity)
It also says in the question that acceleration is uniform (so a constant)
You probably have a formula list somewhere and from that you can figure out the one you need is the one below.
Δd = vi x t + 0.5 x a x t^2 (delta d = total distance moved, vi = initial velocity, a = acceleration, t = time)
a is what we are trying to solve for and delta d is the total change in position so Δd =(-5) - 2.91 =-7.91
plugging everything into the formula we get the following.
-7.91 = 11 x 2.45 + 0.5 x a x 2.45^2
(-7.91-11 x 2.45)/(0.5x2.45^2)=a
a=-11.615cm/s^2

DevoidDecoy [#25]

−11.6 cm/s^2 is what i got but idk if i did the math correct. the idea should be right though.

i used the one of the uniform acceleration equations, displacement = velocity initial time + (1/2) acceleration * time^2 (d = vit + (1/2)at^2). Everything is given to you- the displacement in the x-direction is xf-xi. you have the time interval and initial velocity, so just solve for acceleration.

It's pretty simple with the four main kinematic constant acceleration equations, you just have to get the idea down.
iirc vi*t calculates the intial displacement at the initial velocity over the time interval. But because the object is accelerating, you are gaining more displacement at the increasing speeds over the time interval. Think of it as adding the original displacement of the object if it were traveling at the initial speed (11 cm/s) for 2.45s. But since the object gains velocity from that initial velocity at an acceleration, 1/2at^2 calculates such displacement.

imagine the object traveled at 11cm/s for 2.45s and displaced 26.95 cm. Now imagine we travel at the same speed and displacement right next to the object, treating speed relative. We are both currently traveling at a constant velocity. the object in this scenario gains an imaginary little hand that applies a force and accelerates it. We would begin to see the acceleration because the object gains velocity (in this example we would see the object slow down relative to us because it accelerates in the negative x-direction). That negative displacement from the acceleration is the 1/2at^2 we add to the original displacement at 11cm/s for 2.45s.

S = Vo.t + (1/2)at^2
-7.91 = 11(2.45) + 0.5a(2.45)^2
-34.86 = 0.5a(2.45)^2
-69.72 = 6.0025a
-11.61 cm/s^2= a
yeah pretty accurate

No way that early kinematics 💀 it gets a lot lot worse later on enjoy this while it lasts

Bo_Zenga [#26]

did he edit the linked image what you are doing looks nothing like the question there?

yeah he edited

FireII [#22]

I'm 15 bro i have calculus to do as well 💀

wtf yall have calculus at 15??? thats mental torture no?

CryoZanderDerrekEnjoyer [#31]

wtf yall have calculus at 15??? thats mental torture no?

I live in america, since I lived in India I was super ahead in math so here Im doing calculus at 15. theres a ton of other people my age in my classes too, but idk how it is in India now

FireII [#32]

I live in america, since I lived in India I was super ahead in math so here Im doing calculus at 15. theres a ton of other people my age in my classes too, but idk how it is in India now

stop taking my spot at MIT 😤😤

1phnxe [#33]

stop taking my spot at MIT 😤😤

unfortunately i wear deodorant so i wont be getting into mit 😖

the answer is 69.420

s = ut + .5at^2
-7.91=112.45+.5a2.45^2
-7.91=26.95+3.00125a
-34.86=3.00125a
-11.615=a

FireII [#34]

unfortunately i wear deodorant so i wont be getting into mit 😖

bro wtf 💀💀😭😭😭