Question

(2(43.0)+ 2(3.74)^2/1.62) +- sqrt ((2(43.0) + 2(3.74)^2/3.74))^2 - 4(43.0)^2).....i have solved this thing 7 times now and am making errors somewhere...can someone help me?

Answer 1

it's just the quadratic equation

Answer 2

oh...the entire equation is divided by 2...smh

Answer 3

What's the original problem?

Answer 4

2 trains are on a straight horizontal track. One car starts at rest and is put in motion with a constant acceleration of 1.62 m^2. This car moves toward a car that is 43.0m away and moving at a constant speed of 3.74 m/s. Where will the trains collide?

Answer 5

So for train 1 we have a function: Distance = 1.62t^2
train 2 is Distance = 43 + 3.74t

Answer 6

with a karen added in to make things balance
http://www.wolframalpha.com/input/?i=%282%2843.0%29%2B+2%283.74%29%5E2%2F1.62%29+%2B-+sqrt%28+%28+%282%2843.0%29+%2B+2%283.74%29%5E2%2F3.74%29%29%5E2+-+4%2843.0%29%5E2%29

Answer 7

2 trains are on a straight horizontal track.

One car starts at rest and is put in motion with a constant acceleration of 1.62 m^2.

moves toward a car that is 43.0m away and moving at a constant speed of 3.74 m/s.

Where will the trains collide?

both cars moving?

Answer 8

both cars moving

Answer 9

phi...in the equation only the numerator is multiplied by 2 not the entire fraction

Answer 10

i think their moving toward each other amistre.....i'm not sure...question seems vague

Answer 11

question is very vague in that regards

Answer 12

If you set the two equations I have above equal to eachother and solve you get t = 6.43406 seconds which is at 67.0633 meters

Answer 13

a constant acceleration = 1.62 m^2. per sec? per minute?

Answer 14

i guess b/c it says the trains collide their moving toward each other

Answer 15

I assumed that they were moving in the same direction

Answer 16

says 67.1 is incorrect.....we have to use 3 sig figures

Answer 17

they can be moving in the same direction and the other catch up to it

Answer 18

a(t) = 1.62

v(t) = 1.62 t + C , but at t=0, v(t)=0, so C = 0 if im thinking right

Answer 19

If they are moving at eachother change the middle term in the train 2 to a negative. Setting them equal and solving gives time 4.12542 seconds and distance 27.5709 meters

Answer 20

My equations I'm using are
For train 1 we have a function: Distance = 1.62t^2
Train 2 is Distance = 43 - 3.74t

Answer 21

i took too many tries....answer turned out to be 80.2

Answer 22

i do not see how that happens

Answer 23

A train speed = 1.62 t
position = 1.62 t^2/ 2

B train speed = 3.74 m/s.
position = 3.74 t

right?

Answer 24

with both trains moving in the same direction
http://www.wolframalpha.com/input/?i=solve+0.81+t%5E2+-3.74*t-43%3D0
t= 9.95 seconds
distance will be 0.81*t^2

Answer 25

using d= 0.5 a t^2
and d= 43 + 3.74 t

Answer 26

if the trains move towards each other t= 5.33 seconds
d= 23 m

Answer 27

.81 t^2 = 3.74 t + 40

.81 t^2 -3.74t -40 = 0

t = \(\cfrac{1}{1.62}\left(3.74+\sqrt{3.74^2-4(.81)(-40)}\right)\)
= \(\cfrac{1}{1.62}\left(3.74+\sqrt{13.9876+129.6}\right)\)
= \(\cfrac{1}{1.62}\left(3.74+\sqrt{143.5876}\right)\)
= \(\cfrac{1}{1.62}(15.7228)\)

= \(9.7054321\)
............................
B train position = 3.74 t +40
= 3.74(9.7054321) +40
= 76.298...

which is pretty close to what they say the answer is

Answer 28

prolly accounting for my truncating

Answer 29

lol

Answer 30

i wish i could use the excuse that it's hard...but tis' not....just tedious

Answer 31

use 43 instead of 40

Answer 32

well, and maybe if i used 43 instead of 40 lol

Answer 33

wolfram says the correction to 43 makes t = 9.95169

and 3.74(9.95169) +43 = 80.219 ...