Kinematics-1D

From $t = 0$ to $t = T$, the area under the graph is below the time axis, representing a negative displacement equal to $-\frac{1}{2} v_0 T$.
From $t = T$ to $t = 2T$, the area under the graph is above the time axis, representing a positive displacement equal to $+\frac{1}{2} v_0 T$.
The total displacement is the sum of both areas: $-\frac{1}{2} v_0 T + \frac{1}{2} v_0 T = 0$, making statement A correct.
At time $t = T$, the velocity crosses the time axis from negative to positive values, showing that the particle reverses its direction of motion, making statement B correct.
At $t = 0$, the velocity is $-v_0$ giving a speed of $v_0$, and at $t = 2T$, the velocity is $+v_0$ giving a speed of $v_0$, so initial and final speeds are equal, making statement C correct.
Since all individual statements are correct, statement D is the right choice.

From $t = 0 \text{ s}$ to $t = 5 \text{ s}$, the graph forms a triangle above the time axis with base $5 \text{ s}$ and height $5 \text{ ms}^{-1}$.
Positive area $A_1 = \frac{1}{2} \times 5 \times 5 = 12.5 \text{ m}$.
From $t = 5 \text{ s}$ to $t = 10 \text{ s}$, the graph forms a triangle below the time axis with base $5 \text{ s}$ and height $-5 \text{ ms}^{-1}$.
Negative area $A_2 = \frac{1}{2} \times 5 \times (-5) = -12.5 \text{ m}$.
Total displacement in 10 seconds is $A_1 + A_2 = 12.5 - 12.5 = 0 \text{ m}$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{0}{10} = 0 \text{ ms}^{-1}$.

In graph (a), a closed circle implies that for a single value of time $t$, there are two different values of velocity, and time is also shown moving backwards, which is impossible.
In graph (b), the curve doubles back on itself, meaning two or more velocities exist for the same instant of time $t$, which is physically impossible.
In graph (d), the curve moves backwards in time, which is impossible as time must always increase monotonically.
In graph (c), for every given point on the time axis, there is exactly one unique corresponding velocity value, and time moves continuously forward.
Therefore, graph (c) is the only physically possible velocity-time graph.

In Stage 1 (Acceleration), initial velocity $u_1 = 0 \text{ m/s}$, acceleration $a_1 = 2 \text{ m/s}^2$, time $t_1 = 10\text{ s}$.
Distance covered $s_1 = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} \times 2 \times 10^2 = 100\text{ m}$.
Velocity acquired $v_1 = u_1 + a_1 t_1 = 0 + 2 \times 10 = 20 \text{ m/s}$.
In Stage 2 (Constant Speed), speed $v = 20 \text{ m/s}$, time $t_2 = 30\text{ s}$.
Distance covered $s_2 = v \times t_2 = 20 \times 30 = 600\text{ m}$.
In Stage 3 (Deceleration), initial velocity $u_3 = 20 \text{ m/s}$, final velocity $v_3 = 0 \text{ m/s}$, deceleration $a_3 = -4 \text{ m/s}^2$.
Using $v_3^2 = u_3^2 + 2 a_3 s_3$, we get $0^2 = 20^2 + 2(-4) s_3$, which gives $8 s_3 = 400$, so $s_3 = 50\text{ m}$.
Total distance travelled $s_{\text{total}} = s_1 + s_2 + s_3 = 100 + 600 + 50 = 750\text{ m}$.

In the given $v-t$ graph, the first segment shows velocity decreasing linearly with time, which means the slope $\frac{dv}{dt}$ is negative and constant. Thus, acceleration $a$ is constant and negative.
In the second segment, the velocity remains constant over time (horizontal line), which means the slope $\frac{dv}{dt} = 0$. Thus, acceleration $a = 0$.
In the third segment, the velocity increases linearly with time, which means the slope $\frac{dv}{dt}$ is positive and constant. Thus, acceleration $a$ is constant and positive.
Representing these piecewise constant values gives a step-wise graph where acceleration is initially negative, then zero, and finally positive.
Therefore, graph (b) correctly represents the acceleration versus time graph.

In the given acceleration-time graph, the acceleration is positive and constant for the first interval.
Since $v = \int a \, dt$, a constant positive acceleration means velocity increases at a constant linear rate (straight line with a positive slope).
In the second interval, the acceleration becomes zero.
When $a = 0$, velocity remains constant over time (a flat horizontal straight line).
In the third interval, the acceleration becomes positive and constant again.
This causes velocity to increase linearly once again with a positive slope.
Combining these three regions produces a graph that goes linearly upwards, stays horizontal, and then continues linearly upwards.
Therefore, graph (b) correctly represents the velocity of the body plotted against time.

The displacement of the particle is given as $x = 10 - 4t + 6t^2$.
Differentiating $x$ with respect to time $t$ gives velocity $v = \frac{dx}{dt} = \frac{d}{dt}(10 - 4t + 6t^2) = -4 + 12t$.
At initial time $t = 0 \text{ s}$, the initial velocity is $v(0) = -4 \text{ m/s}$, which is a negative value on the velocity axis.
The slope of the velocity-time graph is the acceleration $a = \frac{dv}{dt} = +12 \text{ m/s}^2$, which is constant and positive.
Therefore, the $v-t$ graph is a straight line starting from a negative value on the velocity axis and increasing with a constant positive slope.
This corresponds to graph (a).