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**+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

#1**+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 4_{kmh-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 120_{km}?

**Average speed:**

^{120 km}/_{1 hour} = 120_{kmh-1}

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

120_{kmh-1} = X-2.

X = 118_{kmh-1}

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

124_{kmh-1} = X

Sir-Emo-Chappington
Aug 23, 2015

#2**+10 **

Best Answer

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