A car slows down from 22m/s to rest in a distance of 82m. What was its acceleration, assumed constant?

Answer 1

Using v^2 = u^2 - 2as, where v is the final velocity, zero in this case, u the initial velocity, 22 m/sec, a the accelertion, and s the distance, then:

(0)^2 = (22)^2 + 2x a x 84, so

0 = 484 + 168a so

168a = -484, so

a = -484/168, so

a = -2.881 m/ sec / sec.

Hope this helps, Twiggy.

Answer 2

Using the formula v^2 = u^2 + 2as, where v=final speed (0 m/s), u=initial speed (22m/s) and s = distance (82m), we get the following:

0^2 = 22^2 + 2 x a x 82

Therefore

0 = 484 + 164 x a

Subtracting 484 from each side, we get

-484 = 164 x a

Dividing by 164, we now see

-484/164 = a

Since 484/164 = c. 2.9512, the acceleration of the vehicle is c. -2.9512 m/s/s.

I hope this helps, but please feel free to drop me a line if you'd like to discuss this or any other figure-work further.

Answer 3

on your e book you will (or damned nicely -ought to-) locate an equation that sounds like this: v² = v0²+2ax the place v is velocity, v0 is preliminary velocity, a is acceleration, and x is distance. you're given v0 (19m/s), you already know the gap (88m), and 'relax' kinda means that v=0. Now grab your calculator, do particularly 7'th grade algebra, and then do the mathematics. in case you don't get a = -2.051 m/s², do it lower back and save doing it till you do. Doug

Answer 4

the answer to this question can be found from the formula 2as=vf²-vi² where ur S value is 82m and vf value is 0 and vi value is 22m/s

Answer 5

use this formula:
Vf^2 = 2ad + Vi^2

Vf = final velocity ( 0m/s)
a = acceleration
d = displacement
Vi = intial velocity

plug in chunks
0^2 = 2(a)(82) + 22^2
-484 = 164a
a =~ -2.95 m/s^2

hope it helps

Answer 6

a=v2-v1/d

so.
a=?
v2=0m/s
v1=22m/s
d=82m
so
(om/s-22m/s)/82
so
a=0.268292683m/ssquared....just round off to the sigfig or nearest 100'th and your good.
so
a=.27

Answer 7

i hate lazy science students who come on yahoo to do their homework.