A train starting from rest first accelerates uniformly up to a speed of $80 \mathrm{~km} / \mathrm{h}$ for time $t$, then it moves with a constant speed for time $3 t$. The average speed of the train for this duration of journey will be (in $\mathrm{km} / \mathrm{h}$) :

<p>To find the average speed of the train for the duration of the journey, we need to know the total distance covered by the train and the total time taken.</p> <p>The train accelerates uniformly to a speed of $80 \, \mathrm{km/h}$ over time $t$, and then moves at this constant speed for $3t$. The average speed can be calculated using the formula:</p> <p>$$\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$$</p> <p><strong>Step 1: Calculate the distance covered during acceleration</strong></p> <p>The distance covered while the train is accelerating can be found using the formula for the distance travelled under uniform acceleration:</p> <p>$d_1 = \frac{1}{2} at^2$</p> <p>Where:</p> <ul> <li>$d_1$ is the distance covered during acceleration.</li> <li>$a$ is the acceleration (we don't have a direct value for this, but we can work with the given information).</li> <li>$t$ is the time.</li> </ul> <p>However, to proceed with the calculation without the acceleration ($a$), we recognize that the formula directly correlates to distance but requires knowledge of acceleration. Instead, let's think in terms of the final speed and time, given that the train reaches $80 \, \mathrm{km/h}$ (or $\frac{80}{3.6} = 22.22 \, \mathrm{m/s}$) in time $t$.</p> <p><strong>Using the relationship between velocity, time, and distance,</strong> since the acceleration is uniform, we can use:</p> <p>$d_1 = v \times t_1 - \frac{1}{2} a t^2$</p> <p>Given that the initial speed $u = 0$ and final speed $v = 80 \, \mathrm{km/h}$, converting the speed to meters per second (since our time is likely in seconds) gives us $22.22 \, \mathrm{m/s}$. But without directly calculating acceleration, we simplify using average speed for the acceleration phase because it starts from rest and reaches $v$.</p> <p>The average speed during acceleration, \(v_{avg} = \frac{u + v}{2} = \frac{0 + 80}{2} = 40 \, \mathrm{km/h}$$.</p> <p></p> <p>Thus, the distance $d_1 = v_{avg} \times t = 40 \, \mathrm{km/h} \times t$.</p> <p><strong>Step 2: Calculate the distance covered at constant speed</strong></p> <p>The distance covered at a constant speed is easier to calculate:</p> <p>$d_2 = v \times t_2 = 80 \, \mathrm{km/h} \times 3t$</p> <p><strong>Step 3: Calculate the total distance and the total time</strong></p> <p>The total distance ($D$) covered is the sum of $d_1$ and $d_2$:</p> <p>$D = d_1 + d_2 = 40t + 240t = 280t \, \mathrm{km}$</p> <p>The total time ($T$) taken is $t + 3t = 4t$.</p> <p><strong>Step 4: Calculate the average speed</strong></p> <p>Substitute the values of $D$ and $T$ in the formula of average speed:</p> <p>$\text{Average speed} = \frac{280t}{4t}$</p> <p>This simplifies to $70 \, \mathrm{km/h}$.</p> <p><strong>So, the average speed of the train for this duration of the journey is $70 \, \mathrm{km/h}$, which matches with Option A.</strong></p>