# The Train was moving at constant speed, then it stopped for $12$ minutes. [closed]

The Train was moving at constant speed, then it stopped for $12$ minutes. After Train started moving again the driver sped up by $20km/h$ and he made up for the lost time in $99km$ distance that was left, what was the speed of the train before it stopped moving?

That's how I tried it,

$99=(v+20)t$

I can't algebraically model this situation correctly, hint appreciated

## closed as off-topic by John B, Leucippus, Shailesh, Xander Henderson, José Carlos SantosJun 16 '18 at 8:14

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• You can't add a speed (measured in km/h) to mulitple of time (in h) and get a distance (in km). But $d = v*t$ and so... – fleablood Jun 15 '18 at 22:02
• Sorry for the bad question, I edited it now, test answer says 90km/h was the trains speed before it stopped moving if it helps. – user567775 Jun 15 '18 at 22:05
• Okay $99 = (v + 20)t$ is the train picking up speed. Can you write an equation that would have happened if the train hadn't stopped and hadn't picked up speed. (If it had been moving for 12 extra minutes.....)? – fleablood Jun 15 '18 at 22:14

$99 = (v+20)t$ is a fine start.

$v$ is the original speed. $v+20$ is the speed after it resumes its travel. $t$ is the time elapsed after resuming travel. And $99 km$ is the distance covered.

But how are you using the information that the train was stopped in the station for 12 minutes?

The average speed, including the 12 minutes wait time equals the original speed.

$\frac {99}{t+\frac {12}{60}} = v$

$\frac {12}{60}$ because we are measuring time in hours and not minutes.

Now you have two equation and two unknowns and can substitue from one equation to the other.

$$99 = vt + 20t\\ 99 = vt + 0.2v$$

Subtracting one from the other we get $v = 100 t$ and substituting. $$100t^2 + 20t - 99 = (10t - 9)(10 t + 11) = 0\\t = 0.9 \text {h},v = 90 \frac {\text {km}}{\text {h}}$$

• okay I found t=(around 4.5), then put it in $99=(v+20)*4.5$ , $99/4.5$ it gives me $22$, and answer is $90km/h$ before it stopped moving, what am I doing wrong? – user567775 Jun 15 '18 at 22:37
• I have lead you astray. The train stops for 12 minutes. But speed is measured in $\frac {\text {km}}{h}.$ We must convert everything into the same set of units. – Doug M Jun 15 '18 at 22:46
• how am I gonna get $90km/h$ though? I don't see any way – user567775 Jun 15 '18 at 22:48
• I have filled in the details. – Doug M Jun 15 '18 at 22:53
• When you subtract one from the other the $vt$ terms cancel and you get $20t - 0.2v = 0.$ Then we solve for $v$ in terms of $t$ (or the other way if you would rather.) – Doug M Jun 15 '18 at 23:03

Let $d_1, t_1, v_1$ be distance, time, and speed of the train before stopping.

Let $d_2, t_2, v_2$ be the distance, time, and speed of the train while it was stopped. (Hint: If the train is stopped then $v_2 = 0$. And $d_2 = .....$?)

Let $d_3, t_3, v_3$ be the distance, time and speed of the train when it picked up speed. (Hint: The question specifically tells you that $d_3 = 99$. It also tells you something about $v_3$...)

Let $d = d_1 + d_2 + d_3$ be total distance and $t=t_1 + t_2 + t_3$ be total time.

The question says the train made up lost time:

That means $d = v_1*t= v_1*t_1 + v_2*t_2 + v_3*t_3$.

Can you set that up?

Can you solve it?

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Since the train stopped it did $99 = (v + 20)t$ to make up for lost time.

But what if it hadn't stopped and had been going at the original speed for the lost $12$ minutes. Then you'd have $99=v(t+12)$.