Physics Problem?

Question

Two trains, one traveling at 64.00 km/h and the other at 128.00 km/h, are headed toward one another along a straight, level track. When they are 736 m apart, each engineer sees the other's train and applies the brakes. What acceleration do the two trains need to have to stop exactly in a distance of 736 m?

Answer 1

x(t)=v0*t-.5*a*t^2
and
v(t)=v0-a*t

if they have the same deceleration magnitude, the stopping times and distances will be different. The stopping distance must be less than 736 m.

x1(t1)=64/3.6*t1-.5*a*t1^2
and, when stopped,
v1(t1)=0=64/3.6-a*t1

for train 2

x2(t2)=128/3.6*t2-.5*a*t2^2
and, when stopped,
v2(t2)=0=128/3.6-a*t2

and
x1(t1)+x2(t2)<=736 m

v1(t1)=0=64/3.6-a*t1
17.78=a*t1
t1=17.78/a
x1(t1)=.5*17.78^2/a
x1(t1)=158/a

35.56=a*t2
t2=35.56/a
x2(t2)=.5*35.56^2/a
x2(t2)=632.1/a

736*a=158+632.1
a=1.0735 m/s^2

check
t1=16.56 seconds
x1=147.2 m

t2=33.12 seconds
x2=588.8 m

Sum=736 m

Check the vs
v1(16.56)=0, good

v2(33.12)=0, good

it checks.

j

Answer 2

736m = .736km

let a be the acceleration of the first train traveling at 64km/hr
let g be the acceleration of the second trian traveling at 128 km/hr

the acceleration of both trains must be large enough so they meet exactly in .736km

Vf = at + Vi

Vf = final velocity
a = acceleration
t = time
Vi = intial velocity

When the two trains stop, their final velocity is 0km/hr

first train
0 = at + 64

second train:
0 = gt + 128


they are initially .736 km apart, when they meet, their distance add to .736 km

Xf = .5at^2 + Vt + Xi

Xf = final position
Xi = intial position

first train:
Xf = .5at^2 + 64t

second train
Xf = .5gt^2 + 128t

first train + second train = .736
.5at^2 + 64t + .5gt^2 + 128t = .736
.5at^2 + .5gt^2 + 192t = .736

from the two equations
0 = at + 64
0 = gt + 128

they both stop in a same amount of time. Solve of t
0 = at + 64
-64 = at
t = -64/a

plug t in
0 = gt + 128
0 = g(-64/a) + 128
-128 = g(-64/a)
g = 128a/64
g = 2a

recall that .5at^2 + .5gt^2 + 192t = .736

and we know g = 2a
.5at^2 + .5(2a)t^2 + 192t = .736
.5at^2 + at^2 + 192t = .736
1.5at^2 + 192t = .736

recal that t = -64/a

1.5a(-64/a)^2 + 192(-64/a) = .736
1.5a(4096/a^2) - 12288/a = .736
6144/a - 12288/a = .736
-6144/a = .736
-6144 = .736a
a =~ -8347.826 km/hr^2

g = 2a
g = 2(-8347.826)
g = -16,695.652 km/hr^2

conver to m/s^2
a =~ -0.644122377 m/s^2 (first train at 64 km/hr)
g =~ -1.28824475 m/s^2 (second train at 128 km/hr)


hope this helps!

Answer 3

With twice the speed the faster train requires 2^2 = 4 times the length of the slower one to stop. So it must stop in 0.8*736 = 588.8 m and the slower train in 0.2*736 = 147.2 m.
a = v^2/(2x)
a = (128/3.6)^2/1177.6 = 1.0735 m/s^2
To check, for the slower train
x = v^2/2a = (64/3.6)^2/2.147 = 147.2 m