Question

10. A train travelling at 72 km/h is checked by track repairs. It retards uniformly for 200 m, covering the best 400 m at constant
speed and accelerates uniformly 172 km in a further 600 m. the time me the constant lower speed is equal to the sum o
the times taken in retardins and accelerating. Med the total time taken
ve​

Answer 1

Answer:

200m @72kmph to v at uniform further speed and gain to 72 kmph in 600m,

we have

v^2- u^2 = 2as and s= ut+/- 1/2 at^2 as formulae for accn and and decn,

t in hrs and s in km v in km/hr f and a km/hr^2

Stage 1

72^2-v^2 =2f*0.2

v=72-f t

72^2- (72-f t)^2 = 0.4f……………….(1)

Middle stage

t’= t+ t’’

vt’= v(t+t’’)=0.4…………………..(2)

Last stage ,

72^2 - v^2= 2 a*0.6

v=72-at’’

72^2-(72-a t’’)^2= 1.2a…………….(3)

v=72-f t=72-at’’

So ft=at’’

As 2a*0.6=2f*0.2

3a= f

t’’=3t

Summing up

72^2-v^2 =2f*0.2

(72+v)(72-v)= 2f*0.2

(72+v)(ft)= 2f*0.2

(72+v)(t)= 0.4

Similarly

(72+v)(t’’)= 1.2

Since for middle region,

v t=v(t+t’’) =0.4

(72+v)( 0.4/v)= 1.6

72+v= 4v

V=24

24(t+t’’)=0.4

t+ t’’=1/60

total time is 1/30 hr= 2 mins

Explanation: