Correct option is (d)

Let's denote the total distance covered by the particle as $2d$, where $d$ is the distance for each half. To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.

For the first half of the journey, the particle covers the distance $d$ at a speed of $6\text{m/s}$. The time taken for this part of the journey can be calculated using the formula $\text{time}=\frac{\text{distance}}{\text{speed}}$. So,

${\text{time}}_1=\frac{d}{6}$

For the second half of the journey, the distance $d$ is further divided into two parts, each covered in equal time intervals. Given the speeds are $9\text{m/s}$ and $15\text{m/s}$ respectively, let's call the equal time intervals $t$. The distances covered in these intervals can be found by $\text{distance}=\text{speed}\times \text{time}$.

For the part covered at $9\text{m/s}$:

$d_1=9t$

For the part covered at $15\text{m/s}$:

$d_2=15t$

Since these two parts together make up the second half of the journey,

$d_1+d_2=d$

$9t+15t=d$

This gives us $24t=d$, and from this, we can find $t=\frac{d}{24}$.

The total time for the second half of the journey is the sum of the times for the two parts, which are equal ($t$ each), so the total time for the second half is $2t$. Since $t=\frac{d}{24}$,

${\text{time}}_2=2\times \frac{d}{24}=\frac{d}{12}$

The total time taken for the entire journey is the sum of the times for the first and second halves:

$\text{total time}={\text{time}}_1+{\text{time}}_2=\frac{d}{6}+\frac{d}{12}$

$\text{total time}=\frac{2d}{12}+\frac{d}{12}=\frac{3d}{12}=\frac{d}{4}$

The total distance is $2d$, and the total time is $\frac{d}{4}$. Therefore, the average speed is calculated as:

$\text{average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2d}{\frac{d}{4}}=\frac{2d}{1}\times \frac{4}{d}=8\text{m/s}$

Thus, the correct answer is Option D: 8 m/s.

Correct option is (a)

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 $80\mathrm{km}/\mathrm{h}$ 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:

$d_1=\frac{1}{2}a{t}^2$

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 $80\mathrm{km}/\mathrm{h}$ (or $\frac{80}{3.6}=22.22\mathrm{m}/\mathrm{s}$) 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 $v=80\mathrm{km}/\mathrm{h}$, converting the speed to meters per second (since our time is likely in seconds) gives us $22.22\mathrm{m}/\mathrm{s}$. 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 $d_1=v_{avg}\times t=40\mathrm{km}/\mathrm{h}\times t$.

Step 2: Calculate the distance covered at constant speed

The distance covered at a constant speed is easier to calculate:

$d_2=v\times t_2=80\mathrm{km}/\mathrm{h}\times 3t$

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 $70\mathrm{km}/\mathrm{h}$.

So, the average speed of the train for this duration of the journey is $70\mathrm{km}/\mathrm{h}$, which matches with Option A.

Correct option is (c)

$\begin{array}{rl}& \mathrm{t}=\alpha {\mathrm{x}}^2+\beta \mathrm{x}\text{ (differentiating wrt time) }\\ & \frac{\mathrm{dt}}{\mathrm{dx}}=2\alpha \mathrm{x}+\beta \\ & \frac{1}{\mathrm{v}}=2\alpha \mathrm{x}+\beta \\ & (\text{ differentiating wrt time) }\\ & -\frac{1}{{\mathrm{v}}^2}\frac{\mathrm{dv}}{\mathrm{dt}}=2\alpha \frac{\mathrm{dx}}{\mathrm{dt}}\\ & \frac{\mathrm{dv}}{\mathrm{dt}}=-2\alpha {\mathrm{v}}^3\end{array}$

Correct option is (c)

$\begin{array}{rl}& x=t^3-6t^2+20t+15\\ & \frac{dx}{dt}=v=3t^2-12t+20\\ & \frac{dv}{dt}=a=6t-12\\ & \text{ When }a=0\\ & 6t-12=0;t=2\sec \\ & \text{ At }t=2\text{ sec }\\ & v=3(2{)}^2-12(2)+20\\ & v=8\mathrm{m}/\mathrm{s}\end{array}$

Correct answer is 3

Given the displacement of the particle $x^2=1+t^2$, we want to find the acceleration, which is the second derivative of displacement with respect to time, $a=\frac{d^{2}x}{d{t}^2}$, and we are given that the acceleration at any time $t$ is $x^{-n}$, for us to find the value of $n$.

First, let's find the first derivative of displacement with respect to time, which gives us the velocity. Differentiating $x^2=1+t^2$ with respect to t, we get:

$2x\frac{dx}{dt}=2t$

This simplifies to:

$\frac{dx}{dt}=\frac{t}{x}$

Now, let's differentiate this velocity to find the acceleration:

$a=\frac{d^{2}x}{d{t}^2}=\frac{d}{dt}\left(\frac{t}{x}\right)$

To differentiate $\frac{t}{x}$ with respect to $t$, we'll use the quotient rule:

$\frac{d}{dt}\left(\frac{t}{x}\right)=\frac{x\cdot 1-t\cdot \frac{dx}{dt}}{x^2}$

Substitute $\frac{dx}{dt}=\frac{t}{x}$ into the equation:

$a=\frac{x(1)-t(\frac{t}{x})}{x^2}=\frac{x-\frac{t^2}{x}}{x^2}=\frac{x^2-t^2}{x^3}$

Recalling that the displacement equation given was $x^2=1+t^2$, substitute this into our expression for acceleration:

$a=\frac{1+t^2-t^2}{x^3}=\frac{1}{x^3}$

Thus, the acceleration of the particle at any time $t$ is $x^{-3}$, which means our value for $n$ is $3$.