Help Needed For Nephew

[kuiperbelt]

The Train

https://ibb.co/G5mthxP

[ZebuSanctuary]

Power = Force x Velocity

Locomotive Force 450 kW = Fv (acting up te slope) --- F = 450kW/v

Resistance Forces (acing down the slope);

80 N/t x 300 = 24,000 N and 6,000 N due to gravity (See link)

https://ibb.co/BcdhTPj

Locomotive Force - Resistance Forces = Resultant Force on Train

450,000/v - 24000 - 300,000 x 9.8/490 = 300,000 x acc --- eq 1

Divide through by 300,000

1.5/v - 0.1 = acc ----- eq 2

Where 'acc' is the acceleration of the train up the slope = v/t (m/s²)

When Vmax occurs the Resultant Force = 0 because acc = 0

Therefore from eq 2

1.5/v - 0.1 = 0

Answer (a.) ---> Vmax = 15 m/s


Part (b.)

When v < Vmax then eq 2 can be written as

dv/dt = 1.5/v - 0.1

multiply both sides by v

v.dv/dt = (1.5 - 0.1v)

v.dv/dt (1.5 - 0.1v) = 1

v.dv/(1.5 - 0.1v) = 1.dt

Now multiply both sides by 1/10 to eliminate decimals;

v.dv/(15 - v) = dt/10

Integrating the equation above would involve integration by parts (messy) therefore simplify and rearrange and reduce to the form;

-[1 - 15/(15 - v)].dv = dt/10 Now perform integral calculus;

-Σ[1 - 15/(15 - v)].dv = 1/10 Σdt

Note due to syntax limitations, Σ represents the elongated S

Integrating both sides yields;

-[v + 15ln(15 - v)] = 0.1t + k --- eq 3

v = 5 when t = 0 (bottom of slope) thus

k = -39.54

now substitute v = 10 into eq 3 to determine t

Answer (b.) t = 53.97 Seconds

Part (c.) Find 'height above sea level' (hasl) when a = 0.125 m/s²

s = ut + at²/2 (distance travelled up the slope in metres)

s = 5 x 53.97 + 0.125 x 53.97²/2

s = 451.9 metres up the slope. By application of basic trigonometry;

hasl = SinΘ x 451.9 ---> 1 x 451.9/490

Answer (c.) = 0.922 metres

or

Approx one yard in old money :-)

[douglas9401]

The illustration suggests that this could be Rocket science. :-)

[kuiperbelt]

Zebu, your answers tally with those of my student neighbour. He sends profuse thanks for providing the solution!! Unfortunately he does not understand how you have rearranged the equation below:

v / (15 - v)

to get

1 - 15/(15 - v)

Could you please elaborate. TIA :-)

[ClareTG0ld]

It's useful to know that you can add zero to any equation, or any part of it, and it will still be the same. So, here, we use v = v + 0 = v - 15 + 15 to get:

v/(15 - v) = (v - 15 + 15)/(15 - v) = (v - 15)/(15 - v) + 15/(15 - v)

and now by magic the first piece is just -1, so that finally

v/(15 - v) = -1 + 15/(15 - v) = -(1 - 15/(15 - v))

which should connect the dots.

Alternatively, you could get your nephew to look up "algebraic long division", but spotting the "add zero trick" is more fun imo.

[kuiperbelt]

Thanks Clare!!

zero = 15 - 15 LOL

Yes, that makes sense to me, so it will most certainly make sense to my student neighbour!!

So simple when you know how LOL