Question

a car is traveling at 16m/s when the brakes are applied. The brakes decelerate the car at the rate of 5.0m/s^2. How long will it take the car to stop?

Answer 1

Use the formula

\[\Large V_{f} = V_{o} + a*t\]

Vf is the final velocity
Vo is the initial velocity
a = acceleration
t = time

Answer 2

In this case

Vf = 0 (we want the car to be stopped)
Vo = 16
a = 5
t = unknown (you're solving for this)

Answer 3

wouldnt it be -5??

Answer 4

because it decelerating?

Answer 5

not time the acceleration

Answer 6

yes sorry, a = -5

Answer 7

ok... so its 0=16+(-5)(t)

Answer 8

correct

Answer 9

-16=-5t

Answer 10

so 3.2=t

Answer 11

getting the same

Answer 12

is it 3.2 seconds

Answer 13

correct

Answer 14

hahaaha

Answer 15

OMG THANK YOU!!!! ok what about this......you want to put a new cd in your car stereo. while driving at 45 mi/h, you look away for 2.0s. What distance (in miles) do you "drive blind"?

Answer 16

all yours @jim_thompson5910 :D

Answer 17

haha :P @skullpatrol

Answer 18

you would use the formula

x = v*t

x = distance traveled
v = velocity
t = time

Answer 19

be sure to convert 2 seconds to hours

or convert 45 mph to mi/sec

Answer 20

uh......what? :/

Answer 21

where are you stuck?

Answer 22

converting that and howd you come up with that equaation?

Answer 23

it's similar to the Distance = Rate * Time equation or D = r*t equation

Answer 24

but with physics, they use x = v*t

same idea, just different variables

Answer 25

how many seconds are in a minute?

Answer 26

ok so d=vt

Answer 27

60

Answer 28

yes that's another way to think of it

Answer 29

and how many minutes in an hour?

Answer 30

60

Answer 31

so there are 60*60 = 3600 seconds in 1 hour

Answer 32

ok yes

Answer 33

1 hour = 3600 seconds

how can you use that to convert 2 seconds to hours?

Answer 34

do 2/3600

Answer 35

0.0006

Answer 36

correct, which is 1/1800 = 0.000555 roughly

Answer 37

a very small piece of an hour

Answer 38

now you can compute d = v*t

Answer 39

v = 45 mph
t = 0.000555 hours

Answer 40

0.025

Answer 41

miles

Answer 42

Great work!

Answer 43

0.024975 ---> 0.025

looks good

Answer 44

yay!!! thanks a bunch!

Answer 45

np

Answer 46

i have one more question if thats ok with you?

Answer 47

go ahead

Answer 48

ok so A typical world-class male sprinter will accelerate to his maximum velocity in 4.0s. If the runner covers a distance of 55m in that time, what is his acceleration?

Answer 49

d=55
t=4
a=?

but what equation should i use?

Answer 50

here's a good chart to have as reference

http://www.studyphysics.ca/newnotes/20/unit01_kinematicsdynamics/chp04_acceleration/images/formulas.GIF

Answer 51

one sec, the formula isn't on that page, but I'll try to find it

Answer 52

haha yea...it would be but i dont know the velocity...

Answer 53

I wonder if the initial velocity is 0?

Answer 54

if so, then we can use the second formula on that link I posted

Answer 55

If initial velocity is 0, then

d = 55
Vi = 0
t = 4

use the second formula to solve for 'a'

Answer 56

but we want to know the acceleration..

Answer 57

ohhh ok

Answer 58

\[\Large d = V_{i}*t + 1/2*at^2\]

\[\Large 55 = 0*4 + 1/2*a*(4)^2\]

solve for 'a'

Answer 59

ok give me just a sec

Answer 60

ok i got 1.72m/s^2

Answer 61

that's too small

Answer 62

uhhh......ok i did 55=0(4)+1/2(a)(4)^2
then 55=1/2(a)(16)
then 27.2=a(16)
then 1.72=a

Answer 63

you made an error going from 55=1/2(a)(16) to 27.2=a(16)

Answer 64

you should multiply both sides by 2, not divide

Answer 65

....wow im retarded

Answer 66

nah you're not, silly mistake really

Answer 67

ok lol gimme a sec...

Answer 68

ok i got 6.9 m/s^2

Answer 69

that just shows that it stoo late to be doing physics haha

Answer 70

I'd go with 6.875

Answer 71

ok thats no problem with me! :)

Answer 72

Thanks a whole bunch! You really helped! I will definitely use that formula table throughout this year! Thanks againn!

Answer 73

you're welcome