Velocity of a train A train travels with a velocity that doubles every hour. In 3 hours, it has travelled 350km. What was the trains velocity in the second hour?
A. 225 km/h
B. 175 km/h
C. 150 km/h
D. 100 km/h
Answer is D. Can someone explain?
 A: Let's say the train starts with some initial velocity $v_0$, which we assume to have units of kilometers per hour. First it travels one hour at velocity $v_0$, then one hour at velocity $2v_0$ and finally one hour at $4v_0$. Since we know the total distance traveled after 3 hours, we can find $v_0$ by solving:
$$v_0+2v_0+4v_0=350$$
So $v_0=\frac{350}{7}=50$. After one hour, the train has doubled its velocity once, so during the second hour, the train travels at $100 \frac{km}{h}$.  
A: Let $x$ be the train's velocity the first hour, $y$ in the second hour,$z$ in the third.
We have the following relations:
$x + y + z = 350$ and $2x=y$ and $4x=z$
So we have $x + 2x +4x =350 km$
$$7x=350$$
hence
$x=50$ and $y=100$
A: It is interesting to note that, whether you assume
(i) that the velocity remains constant for any hour, or
(ii) that the velocity increases exponentially (and "continuously") such that it doubles every hour,
$\color{red}{\text{the answer is the same!}}$
Several answers for (i) have already given above. 
The solution below is for (ii).
First we note that 
$$\int2^t\; dt=\int e^{t\ln 2}\; dt=\frac {e^{t\ln 2}}{\ln 2}+c=\frac {2^t}{\ln 2}+c$$
Assume the velocity at $t=0$ is $\lambda$. As velocity doubles each hour, this means that 
$$V(t)=\lambda\cdot 2^t\qquad\text{($t$ in hours)}$$
Distance in the first three hours is $350$km (given), i.e. 
$$\begin{align}
350&=\int_0^3 V(t) dt =\lambda \int_0^3 2^t\; dt=\lambda \left[\frac {2^t}{\ln 2}\right]_0^3=\frac{7\lambda}{\ln 2}&&\cdots(1)
\end{align}$$
Average velocity during the second hour is given by
$$\begin{align}\overline{V}_2&=\frac{\text{Distance travelled}}{\text{Time taken}}=\frac {\int_1^2V(t)}1\; dt=\lambda\int_1^2 2^t\; dt=\lambda\left[\frac {2^t}{\ln 2}\right]_1^2=\frac {2\lambda}{\ln 2}&&\qquad \cdots (2) 
\end{align} $$
From $(1)$ and $(2)$, 
$$\overline{V}_2=\frac 27\cdot 350=100[\text{km/h}]\quad\blacksquare$$
A: Assuming that the intent is to solve this using linear algebra, here are the steps
$$ x =\mbox{distance traveled in the first hour}$$
$$ y =\mbox{distance traveled in the second hour}$$
$$ z =\mbox{distance traveled in the third hour}$$
This problem is expressed mathematically as
$$x+y+z=350$$
$$2x=y$$
$$2y=z$$
Which can be rewritten as
$$x+y+z=350$$
$$2x-y+0z=0$$
$$0x+2y-z=0$$
The augmented matrix of this linear system is
$$\left[\begin{array}{ccc|c}
1 & 1 & 1 & 350 \\
2 & -1 & 0 & 0 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 1 & 1 & 350 \\
0 & -3 & -2 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & -3 & -2 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & -7 & 0 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & 1 & 0 & 100 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 0 & 0 & 50 \\
0 & 1 & 0 & 100 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 0 & 0 & 50 \\
0 & 1 & 0 & 100 \\
0 & 0 & 1 & 200 \\
\end{array}\right]$$
Which implies that
$$ x =50\ \mathrm{km}$$
$$ y =100\ \mathrm{km}$$
$$ z =200\ \mathrm{km}$$
Since $y$ represents the distance that the train traveled in the second hour, the answer then is $100 \frac{\mathrm{km}}{h}$.
