Question 16: Motion in 1D

Question

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 $$\mathrm{km} / \mathrm{h}$$) :

Accepted Answer

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.
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:
$$\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$$
Step 1: Calculate the distance covered during acceleration
The distance covered while the train is accelerating can be found using the formula for the distance travelled under uniform acceleration:
$$d_1 = \frac{1}{2} at^2$$
Where:
$d_1$ is the distance covered during acceleration.
$a$ is the acceleration (we don't have a direct value for this, but we can work with the given information).
$t$ is the time.
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 $$t$$.
Using the relationship between velocity, time, and distance,
since the acceleration is uniform, we can use:
$$d_1 = v \times t_1 - \frac{1}{2} a t^2$$
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 $v$.
The average speed during acceleration, \(v_{avg} = \frac{u + v}{2} = \frac{0 + 80}{2} = 40 \, \mathrm{km/h}$$.
Thus, the distance $d_1 = v_{avg} \times t = 40 \, \mathrm{km/h} \times t$.
Step 2: Calculate the distance covered at constant speed
The distance covered at a constant speed is easier to calculate:
$$d_2 = v \times t_2 = 80 \, \mathrm{km/h} \times 3t$$
Step 3: Calculate the total distance and the total time
The total distance ($D$) covered is the sum of $d_1$ and $d_2$:
$$D = d_1 + d_2 = 40t + 240t = 280t \, \mathrm{km}$$
The total time ($T$) taken is $t + 3t = 4t$.
Step 4: Calculate the average speed
Substitute the values of $D$ and $T$ in the formula of average speed:
$$\text{Average speed} = \frac{280t}{4t}$$
This simplifies to $70 \, \mathrm{km/h}$.
So, the average speed of the train for this duration of the journey is $70 \, \mathrm{km/h}$, which matches with Option A.

Answer

70

Answer

40

Answer

30

Answer

80

Motion in 1D

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

Choose the correct answer: