Question

A car starts from rest and accelerates at a rate of 3.00 m/s^2 for 10 seconds and then maintains a constant speed for 25 seconds. The car then decelerates at -2.50 m/s^2 until it stops. Find the total distance covered.

Answer 1

its too hard :( im looking for examples

Answer 2

PHYSICS!!!!!

Answer 3

WHY DOES NOBODY HELP ME

Answer 4

Lol warood try giving dominique some help if mines to hard

Answer 5

THANK YOU SO MUCH DWADE

Answer 6

BTW DWAYNE WADE IS A FRKIN BEAST

Answer 7

HELL YEA! i hate the heat though just wade is cool

Answer 8

wade has always been cool

Answer 9

It's three parts, right?

Answer 10

D = d1+d2+d3

Answer 11

Total Distance = DT
DT = D1 + D2 + D3 (D1/2/3 are different parts of acceleration)

D1 = vi*t + 1/2at^2
D1 = 0 + .5*(3m/s^2)*10^2
D1 = figure it out ;-)

D2 = vi*t + 1/2at^2
vi of D2 = 3m/s^2 * 10s = 30m/s
D2 = 30m/s*25 + 0 (acceleration = 0)

D3 = vi*t +1/2at^2
D3 = 30*t + 1/2*(-2.5m/s^2)*t^2
t = 30/2.5 = 12
D3 = 30*12 + 1/2 (-2.5m/s^2)*30^2

Plug 'n chug

Answer 12

so Vi is the initial velocity?

Answer 13

Vi is the initial velocity *in that section*

So Vi in the first part is different from Vi in the 2nd and 3rd part.

Answer 14

yah... sorry for not delineating that. I have never typed physics before :D

Answer 15

yea its weird its not your fault :P

Answer 16

:) no worries

Answer 17

so let me get this straight cuz im an idiot when it comes to physics, for the d1 i got 59 is that remotely correct?

Answer 18

I don't think so. What did you use for

vi = ?
t = ?
a = ?

Answer 19

good luck guys... gotta go write a paper -.-

Answer 20

for vi i used 3.00 for t i used 10 and for a used .5?

Answer 21

alright @micah.daniel.anderson thanks for the help

Answer 22

"A car starts from rest and accelerates at a rate of 3.00 m/s^2 for 10 seconds "

\[d_1 = v_{i}^2 + \frac{1}{2}at^2\]

So how fast is the car going in the beginning?

Answer 23

does that say v^2?

Answer 24

would it be 0+1/5*3.00*10^2?

Answer 25

i mean 0+1/2*3.00*10^2

Answer 26

is the car moving at 45 m/s in the beginning?

Answer 27

yeah!

Answer 28

oh
, nononon

Answer 29

you were right before, with Vi = 0

Answer 30

and that's (Vi)^2, yeah

Answer 31

oh alright

Answer 32

and so a = 3.00 and t = 10 correct?

Answer 33

yes yes :)

Answer 34

so then the equation would read 0^2+0.5*3.00*10^2?

Answer 35

yup yup - just remember all your units if you're writing it on paper :) (they get messy when you type em, I know :P)

Answer 36

ill keep that in mind lol. so the outcome gives me 150 thats meters per second?

Answer 37

or is that just the distance covered initially?

Answer 38

You used this equation, right?

\[d = v_i t + \frac{1}{2}at^2\]

*distance* is equal to velocity times time plus....

so units of distance in this case are meters, m

It's easy to remember units when you say what the terms mean as opposed to just letters :)

And yeah, that's the distance covered in this section: