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

Question

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

Answer 1

$99 = (v+20)t$ 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. $$100t^2 + 20t - 99 = (10t - 9)(10 t + 11) = 0\\t = 0.9 \text {h},v = 90 \frac {\text {km}}{\text {h}}$$

Answer 2

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 $99=v(t+12)$.