A train traveling at 𝑣o=27.0 m/s begins to brake by applying a velocity‑dependent instantaneous acceleration 𝑎(𝑣)=𝛼/(𝑣+𝑢) m/s^2?
A train traveling at 𝑣o=27.0 m/s begins to brake by applying a velocity‑dependent instantaneous acceleration
𝑎(𝑣)=𝛼/(𝑣+𝑢) m/s^2
where 𝛼=−20.0 m2/s3, 𝑣 is the instantaneous velocity of the train, and 𝑢=0.5 m/s.
Determine the distance 𝐷 traveled by the train before it comes to a complete stop.
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Verified answer
a(v)=-20/(v+0.5)
=>
v(dv/dD)=-20/(v+0.5)
=>
v(v+0.5)dv= -20dD
=>
(v^3)/3+(v^2)/4=-20D+C
[after integrations]
D=0 & v=27
=>
C=6743.25
=>
(v^3)/3+(v^2)/4=-20D+
6743.25
When the train stops, v=0
=>
20D=6743.25
D=337.163 m approximately.
dv/dt = -20/(v + 0.5) =>
(v + 0.5) dv = -20 dt =>
(1/2)v^2 + 0.5v = -20t + C.
The "C" must be (1/2)(27^2 + 27) = 378, so
(1/2)(v^2 + v) = 378 - 20t.
If v = 0 then t = 18.9 seconds.
Next, you need x(t), where
d2x/dt2 + dx/dt = 2(378 - 20t) = 756 - 40t.
This diff eq is not separable, but if you Google "pauls linear differential equation" and select the item dated June 3, 2018, you can see how to solve it.