A train starts from $X$ towards $Y$, which is at a distance of $55$ km, at a speed of $40$ km per hour. After running a certain distance, it increases its speed to $50$ km per hour and reaches $Y$ in $1$ hour and $15$ minutes after leaving $X$. After what time did the train change its speed?
My attempt: $$x/40+(55-x)/50=75$$
Best Answer
As the train is moving at a constant speed until it changes to a different constant speed, its journey is governed by the equation,
$$s = v_1t_1 + v_2t_2$$ where $s$ is distance traveled, $v_1$ and $t_1$ are speed and time of the first half, and $v_2$ and $t_2$ are the same for the second half.
We know that the total time taken was 1 hour and 15 minutes, or 1.25 hours. We know that up until time $t_1$ the train is moving at 40 km/h, and afterwards it is moving at 50km/h. Therefore we have $s = 55$, $v_1 = 40$, $v_2 = 50$, $t_2 = 1.25 - t_1$.
Hence putting these values into the above equation, we can solve for $t_1$, the time at which the train changes speed. $$40t_1 + 50(1.25-t_1) = 55$$ $$\implies 62.5 - 10t_1 = 55$$ $$\implies t_1 = 0.75 \text{ hours}$$ $$\implies t_1 = 45 \text{ minutes}$$