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a train leaves a station 1 hour before the shedules time .the driver decreases the speed by 4km/h.at next station 120km away,the train reached on scheduled time .the original speed of the train is (in km/h)

matsunnymat  Aug 23, 2015

#2
+85786
+10

I'm assuming that the train travels at its normal rate for the first 120 km and then reduces its rate by 4km/hr for the rest of the trip.......

There is no specific answer for this.......to see why...let the normal rate be = R, the normal time = T, and the distance = D...and we have....

RT = D  →  T = D/R

So the time for the first part of the trip = D/R  = 120/R

And the fime for the second part of the trip = [D - 120] / (R - 4]

And the sum of these  equal to the normal time plus 1 hour.....so we have.....

120/R  + [D - 120] / [R - 4]  = T + 1

120/R  + [D - 120] / [R - 4] =  D/R + 1       simplify

120[R - 4] + R[D - 120]  = (R^2 - 4R] [D/R + 1]

120R - 480  +  RD - 120R  = RD + R^2 - 4D - 4R     rearrange

R^2 - 4R - [4D - 480] = 0  ......   now let D = 480km     and we have

R^2 - 4R - [4(480) - 480] = 0

R^2 - 4R  - 1440  = 0    factor

(R + 36) (R - 40)  = 0     .......  and R = 40km/hr

So...when D = 480km, R = 40km/hr and the total time = 480/40 + 1 = 13 hrs

Now,  let D = 960km   and we have

R^2 - 4R - [4(960) - 480] = 0

R^2 - 4R - 3360 = 0      factor

(R -60)(R + 56) = 0      and R = 60km/hr

So, when D = 960km, R = 60km/hr, and the total time = 960/60 + 1 = 17 hrs

Thus......there are infinite possibilities........

CPhill  Aug 24, 2015
Sort:

#1
+423
+5

The wording is a bit confusing.

If I understand correctly, you mean to say the train takes 1 hour to travel 120 kilometres. It starts at X speed and slows down by 4kmh-1.

However, do you mean the speed reduction happened before it left the first station, or are we suggesting it gradually slowed down by this amount uniformly across the 120km?

Average speed:

120 km/1 hour = 120kmh-1

If we suggest it slowed down uniformly across the 120km, then 120kmh-1 = X + (X-4) / 2

120kmh-1 = X-2.

X = 118kmh-1

If we suggest is slowed down before it even started the 120kmh-1 stretch, then 120kmh-1 = X-4

124kmh-1 = X

Sir-Emo-Chappington  Aug 23, 2015
#2
+85786
+10

I'm assuming that the train travels at its normal rate for the first 120 km and then reduces its rate by 4km/hr for the rest of the trip.......

There is no specific answer for this.......to see why...let the normal rate be = R, the normal time = T, and the distance = D...and we have....

RT = D  →  T = D/R

So the time for the first part of the trip = D/R  = 120/R

And the fime for the second part of the trip = [D - 120] / (R - 4]

And the sum of these  equal to the normal time plus 1 hour.....so we have.....

120/R  + [D - 120] / [R - 4]  = T + 1

120/R  + [D - 120] / [R - 4] =  D/R + 1       simplify

120[R - 4] + R[D - 120]  = (R^2 - 4R] [D/R + 1]

120R - 480  +  RD - 120R  = RD + R^2 - 4D - 4R     rearrange

R^2 - 4R - [4D - 480] = 0  ......   now let D = 480km     and we have

R^2 - 4R - [4(480) - 480] = 0

R^2 - 4R  - 1440  = 0    factor

(R + 36) (R - 40)  = 0     .......  and R = 40km/hr

So...when D = 480km, R = 40km/hr and the total time = 480/40 + 1 = 13 hrs

Now,  let D = 960km   and we have

R^2 - 4R - [4(960) - 480] = 0

R^2 - 4R - 3360 = 0      factor

(R -60)(R + 56) = 0      and R = 60km/hr

So, when D = 960km, R = 60km/hr, and the total time = 960/60 + 1 = 17 hrs

Thus......there are infinite possibilities........

CPhill  Aug 24, 2015
#3
+92221
0

Thanks Sir-Emo-Chappington and CPhill,

This question looks really interesting :))

Melody  Aug 24, 2015

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