Instantaneous Velocity and Speed and Velocity-time Graph Questions in English

(D) The displacement is the algebraic sum of the areas under the velocity-time graph.
Area $1$ ($0$ to $2\,s$): $2 \times 8 = 16\,m$
Area $2$ ($2$ to $4\,s$): $2 \times (-4) = -8\,m$
Area $3$ ($4$ to $8\,s$): $4 \times 4 = 16\,m$
Area $4$ ($8$ to $10\,s$): $2 \times (-4) = -8\,m$
Displacement $= 16 - 8 + 16 - 8 = 16\,m$
Distance is the sum of the magnitudes of the areas.
Distance $= |16| + |-8| + |16| + |-8| = 16 + 8 + 16 + 8 = 48\,m$
Ratio of displacement to distance $= \frac{16}{48} = \frac{1}{3}$ or $1: 3$.

(A) The position of the particle is given by $x = 4t^2$.
To find the velocity $v$,we differentiate the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(4t^2) = 8t$.
Now,substitute the given time $t = 5 \, s$ into the velocity equation:
$v = 8 \times 5 = 40 \, ms^{-1}$.
Therefore,the velocity of the particle at $t = 5 \, s$ is $40 \, ms^{-1}$.

(A) The velocity $v$ is given by the slope of the $x-t$ graph,i.e.,$v = \frac{dx}{dt}$.
$(A)$ The $x-t$ graph is a parabola opening upwards,indicating increasing slope. Thus,$v$ increases with time. This matches graph $II$.
$(B)$ The $x-t$ graph shows a decreasing position with a decreasing slope (magnitude),approaching zero. This corresponds to a negative velocity whose magnitude decreases towards zero. This matches graph $IV$.
$(C)$ The $x-t$ graph shows a constant positive slope followed by a constant negative slope. This corresponds to a positive constant velocity followed by a negative constant velocity. This matches graph $III$.
$(D)$ The $x-t$ graph is a straight line with a constant positive slope,indicating constant positive velocity. This matches graph $I$.
Therefore,the correct matching is $(A)-(II), (B)-(IV), (C)-(III), (D)-(I)$.

(A) For an object moving with constant positive acceleration $(a > 0)$,the position-time equation is given by the kinematic equation: $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$.
If we assume the initial position $x_0 = 0$ and initial velocity $v_0 = 0$,the equation simplifies to $x = \frac{1}{2} a t^2$.
This equation represents a parabola opening upwards in the $x-t$ plane.
As $t$ increases,the slope of the tangent to the curve (which represents velocity $v = \frac{dx}{dt} = at$) also increases,indicating that the velocity is increasing over time,which is the definition of positive acceleration.
Therefore,the graph shown in option $A$ represents motion with positive acceleration.

(C) The area under a velocity-time graph represents the displacement of the body,not necessarily the distance (distance is the area under the speed-time graph). Therefore,Statement $I$ is technically false because it specifies 'distance' instead of 'displacement'.
The area under an acceleration-time graph is given by $\int a \, dt = \int \frac{dv}{dt} \, dt = \Delta v$. This represents the change in velocity. Therefore,Statement $II$ is true.
Thus,Statement $I$ is false and Statement $II$ is true.

(A) $1$. The position-time graph shows the distance from the school on the $x$-axis and time on the $t$-axis. The school is at the origin $(x=0)$.
$2$. The point where the graph intersects the $t$-axis represents the time when the student starts their journey. Since $B$ starts later than $A$,and the graph for $A$ reaches the $x$-axis intercept (home) at a lower $x$ value compared to $B$ at the same time,we analyze the intercepts.
$3$. Looking at the graph,the $x$-intercept for $A$ is lower than the $x$-intercept for $B$. Thus,$A$ lives closer to the school. Statement $(A)$ is correct.
$4$. The slope of the position-time graph represents velocity $(v = dx/dt)$. Since the slope of line $B$ is greater than the slope of line $A$,$B$ travels faster than $A$. Statement $(E)$ is correct.
$5$. Therefore,statements $(A)$ and $(E)$ are correct.

(C) The distance is the total area under the $v-t$ graph (taking all areas as positive),and displacement is the net area (taking areas below the time axis as negative).
$1$. Area from $t=0$ to $t=5\,s$ (triangle): $\frac{1}{2} \times 5 \times 10 = 25\,m$.
$2$. Area from $t=5$ to $t=10\,s$ (rectangle): $5 \times 10 = 50\,m$.
$3$. Area from $t=10$ to $t=15\,s$ (trapezium): $\frac{1}{2} \times (10 + 20) \times 5 = 75\,m$.
$4$. Area from $t=15$ to $t=20\,s$ (triangle): $\frac{1}{2} \times 5 \times 20 = 50\,m$.
$5$. Area from $t=20$ to $t=25\,s$ (triangle below axis): $\frac{1}{2} \times 5 \times (-20) = -50\,m$.
Total distance $= 25 + 50 + 75 + 50 + |-50| = 250\,m$.
Total displacement $= 25 + 50 + 75 + 50 - 50 = 150\,m$.
Ratio of distance to displacement $= \frac{250}{150} = \frac{5}{3}$.

(D) The distance $s$ is given as a function of time $t$ by the equation $s = 2.5 t^2$.
The instantaneous speed $v$ is defined as the time derivative of the distance $s$,given by $v = \frac{ds}{dt}$.
Substituting the expression for $s$: $v = \frac{d}{dt}(2.5 t^2)$.
Using the power rule for differentiation,$\frac{d}{dt}(t^n) = n t^{n-1}$,we get $v = 2.5 \times 2 \times t = 5t$.
To find the instantaneous speed at $t = 5\,s$,substitute $t = 5$ into the expression for $v$:
$v = 5 \times 5 = 25\,m/s$.

(C) The position of the particle is given by the equation $x = 5t^2 - 4t + 5$.
To find the velocity $v$,we differentiate the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(5t^2 - 4t + 5)$.
Applying the power rule,we get $v = 10t - 4$.
Now,substitute $t = 2 \, s$ into the velocity equation:
$v = 10(2) - 4 = 20 - 4 = 16 \, ms^{-1}$.
Therefore,the magnitude of the velocity at $t = 2 \, s$ is $16 \, ms^{-1}$.

(D) The distance covered is equal to the area under the velocity-time graph.
For $t = 0$ to $t = 2 \ s$,the graph is a trapezoid with parallel sides $200 \ m/s$ and $400 \ m/s$,and height $2 \ s$.
Area $1 = \frac{1}{2} \times (200 + 400) \times 2 = 600 \ m$.
For $t = 2 \ s$ to $t = 30.5 \ s$,the velocity is constant at $400 \ m/s$.
Time interval $\Delta t = 30.5 - 2 = 28.5 \ s$.
Area $2 = 400 \times 28.5 = 11400 \ m$.
Total distance $= 600 + 11400 = 12000 \ m$.
Converting to kilometers: $12000 \ m = 12 \ km$.

(A) The distance covered by an object is equal to the area under the velocity-time graph.
To find the distance between $t = 0 \; s$ and $t = 4 \; s$,we calculate the area of the region under the graph from $t = 0$ to $t = 4$.
This region consists of a triangle from $t = 0$ to $t = 2$ and a rectangle from $t = 2$ to $t = 4$.
Area of triangle (from $t = 0$ to $t = 2$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \; s \times 10 \; m/s = 10 \; m$.
Area of rectangle (from $t = 2$ to $t = 4$) $= \text{length} \times \text{width} = (4 - 2) \; s \times 10 \; m/s = 2 \; s \times 10 \; m/s = 20 \; m$.
Total distance $= 10 \; m + 20 \; m = 30 \; m$.

(D) The average velocity is given by $\langle v \rangle = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$. The instantaneous velocity is the slope of the $x-t$ graph,$v = \frac{dx}{dt}$.
$(A)$ From $t = 0$ to $t = 3\ s$: $x_i = 0$,$x_f = 5$. $\langle v \rangle = \frac{5 - 0}{3 - 0} = \frac{5}{3}\ m/s$. Statement $(A)$ is incorrect.
$(B)$ From $t = 3$ to $t = 5\ s$: $x_i = 5$,$x_f = 5$. $\langle v \rangle = \frac{5 - 5}{5 - 3} = 0\ m/s$. Statement $(B)$ is correct.
$(C)$ At $t = 2\ s$,the graph is a straight line passing through $(0, -5)$ and $(2, 0)$. Slope $= \frac{0 - (-5)}{2 - 0} = 2.5\ m/s$. Statement $(C)$ is incorrect.
$(D)$ From $t = 5$ to $t = 7\ s$: $x_i = 5$,$x_f = 0$. $\langle v \rangle = \frac{0 - 5}{7 - 5} = -2.5\ m/s$. At $t = 6.5\ s$,the slope is $\frac{0 - 10}{7 - 6} = -10\ m/s$. Statement $(D)$ is incorrect.
$(E)$ From $t = 0$ to $t = 9\ s$: $x_i = 0$,$x_f = 0$. $\langle v \rangle = \frac{0 - 0}{9 - 0} = 0\ m/s$. Statement $(E)$ is correct.
Wait,re-evaluating the options based on the graph:
$(B)$ is correct $(0\ m/s)$.
$(E)$ is correct $(0\ m/s)$.
Looking at the provided options,$(B), (C), (E)$ is option $(D)$. Let's re-check $(C)$: At $t=2$,the line passes through $(1, -5)$ and $(3, 5)$. Slope $= \frac{5 - (-5)}{3 - 1} = \frac{10}{2} = 5\ m/s$. So $(C)$ is correct.
Thus,$(B), (C), (E)$ are correct.

(A) The average velocity is given by the total displacement divided by the total time. The displacement is equal to the area under the velocity-time graph.
Let $V_{max}$ be the maximum velocity at time $t_1$. From the graph,we have:
$V_{max} = t_1 \tan(30^\circ) = t_2 \tan(60^\circ)$
$V_{max} = t_1 \left(\frac{1}{\sqrt{3}}\right) = t_2 (\sqrt{3})$
Therefore,$\frac{t_1}{t_2} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3$,which means $t_1 = 3t_2$.
The displacement during time $t_1$ is $S_1 = \frac{1}{2} \times t_1 \times V_{max}$.
The average velocity during $t_1$ is $V_{avg1} = \frac{S_1}{t_1} = \frac{1}{2} V_{max}$.
The displacement during time $t_2$ is $S_2 = \frac{1}{2} \times t_2 \times V_{max}$.
The average velocity during $t_2$ is $V_{avg2} = \frac{S_2}{t_2} = \frac{1}{2} V_{max}$.
Thus,the ratio of the average velocities is $\frac{V_{avg1}}{V_{avg2}} = \frac{\frac{1}{2} V_{max}}{\frac{1}{2} V_{max}} = 1 : 1$.

(A) From the given $v-x$ graph,the equation of the straight line is $v = mx + c$.
Here,the intercept $c = 20$ and the slope $m = \frac{0 - 20}{10 - 0} = -2$.
So,the equation is $v = -2x + 20$.
We know that acceleration $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx} = \frac{d}{dx}(-2x + 20) = -2$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (-2x + 20)(-2) = 4x - 40$.
This is the equation of a straight line with a slope of $4$ and a y-intercept of $-40$.
At $x = 0$,$a = -40$.
At $x = 10$,$a = 4(10) - 40 = 0$.
Thus,the graph represents a straight line starting from $a = -40$ at $x = 0$ and reaching $a = 0$ at $x = 10$.
Therefore,option $A$ is correct.

(C) The acceleration of a moving body is defined as the rate of change of velocity with respect to time.
Mathematically,$a = \frac{dv}{dt}$.
In a velocity-time graph,the slope is defined as $\frac{\Delta v}{\Delta t}$,which represents the instantaneous or average acceleration.
Therefore,the slope of the velocity-time graph gives the acceleration of the body.

(D) The distance covered by a body in a velocity-time graph is equal to the area under the curve between the given time intervals.
Step $1$: Identify the shape of the graph from $t = 6 \ s$ to $t = 9 \ s$.
At $t = 6 \ s$,$v = 20 \ m/s$. At $t = 9 \ s$,the velocity is $30 \ m/s$ (since the graph is a straight line from $(6, 20)$ to $(10, 40)$,the slope is $m = (40-20)/(10-6) = 5$. Thus,at $t = 9 \ s$,$v = 20 + 5(9-6) = 35 \ m/s$).
Step $2$: Calculate the area of the trapezoid formed between $t = 6 \ s$ and $t = 9 \ s$.
The parallel sides are $v_1 = 20 \ m/s$ and $v_2 = 35 \ m/s$. The height (time interval) is $h = 9 - 6 = 3 \ s$.
Area $= \frac{1}{2} \times (v_1 + v_2) \times h = \frac{1}{2} \times (20 + 35) \times 3 = \frac{1}{2} \times 55 \times 3 = 82.5 \ m$.

(A) In the given displacement-time graph,the position $x$ of the object remains constant as time $t$ increases.
Since the slope of the displacement-time graph represents velocity $(v = dx/dt)$,and the slope of a horizontal line is zero,the velocity of the object is zero.
Therefore,the graph indicates that the object is at rest.

(B) The distance covered is equal to the area enclosed by the velocity-time graph with the time axis.
Total distance covered in the interval from $t = 1 \ s$ to $t = 7 \ s$ (as per the graph,the motion ends at $t = 7 \ s$):
The area is a trapezoid with parallel sides of length $(5-3) = 2 \ s$ and $(7-1) = 6 \ s$,and height $10 \ m/s$.
Total distance $S = \frac{1}{2} \times (2 + 6) \times 10 = 40 \ m$.
Distance covered in the last two seconds (from $t = 5 \ s$ to $t = 7 \ s$):
This is a triangle with base $(7-5) = 2 \ s$ and height $10 \ m/s$.
$S_1 = \frac{1}{2} \times 2 \times 10 = 10 \ m$.
Therefore,the ratio is $\frac{S_1}{S} = \frac{10}{40} = \frac{1}{4}$.

(C) Displacement is the area under the velocity-time graph considering the sign of velocity. Distance is the area under the velocity-time graph taking the absolute value of velocity.
For $t = 0$ to $2 \ s$: Velocity $v = 3 \ m/s$. Area $= 3 \times 2 = 6 \ m$.
For $t = 2$ to $3 \ s$: Velocity $v = -2 \ m/s$. Area $= -2 \times 1 = -2 \ m$.
For $t = 3$ to $5 \ s$: Velocity $v = 2 \ m/s$. Area $= 2 \times 2 = 4 \ m$.
For $t = 5$ to $6 \ s$: Velocity $v = -2 \ m/s$. Area $= -2 \times 1 = -2 \ m$.
For $t = 6$ to $8 \ s$: Velocity $v = 2 \ m/s$. Area $= 2 \times 2 = 4 \ m$.
Total Displacement $= 6 - 2 + 4 - 2 + 4 = 10 \ m$.
Total Distance $= |6| + |-2| + |4| + |-2| + |4| = 6 + 2 + 4 + 2 + 4 = 18 \ m$.
Ratio of displacement to distance $= 10 / 18 = 5 / 9$.

(A) The velocity of an object is the time derivative of its position function,$V(t) = \frac{dR}{dt}$.
For car $A$: $V_{A}(t) = \frac{d}{dt}(at + bt^2) = a + 2bt$.
For car $B$: $V_{B}(t) = \frac{d}{dt}(xt - t^2) = x - 2t$.
To find the time $t$ when the cars have the same velocity,we set $V_{A} = V_{B}$:
$a + 2bt = x - 2t$
Rearranging the terms to solve for $t$:
$2bt + 2t = x - a$
$t(2b + 2) = x - a$
$t = \frac{x - a}{2(b + 1)}$.

(B) The acceleration of a body is given by the slope of its velocity-time graph.
Acceleration $a = \frac{dv}{dt} = \tan \theta$,where $\theta$ is the angle the line makes with the time axis.
For body $A$,the angle with the time axis is $25^{\circ}$. Thus,$a_A = \tan 25^{\circ}$.
For body $B$,the angle with the time axis is $50^{\circ}$. Thus,$a_B = \tan 50^{\circ}$.
Therefore,the ratio of the acceleration of $A$ to the acceleration of $B$ is $\frac{a_A}{a_B} = \frac{\tan 25^{\circ}}{\tan 50^{\circ}}$.

(B) The displacement of an object is given by the area under the $v-t$ graph.
$s = \text{Area under } v-t \text{ graph}$
In the given graph, the area under the uniform motion part (a rectangle) is greater than the area under the uniform acceleration part (a trapezoid) for the same time interval.
Therefore, the displacement during uniform acceleration is less than that during uniform motion.

(A) The given graph is a straight line with a positive intercept on the $v$-axis and a negative slope.
Its equation can be written as:
$v = -mx + v_0 \dots(1)$
where $m = \tan \theta = \frac{v_0}{x_0}$ is the magnitude of the slope.
The acceleration $a$ is given by $a = v \frac{dv}{dx}$.
From equation $(1)$,differentiating $v$ with respect to $x$ gives:
$\frac{dv}{dx} = -m$
Substituting this into the expression for acceleration:
$a = v(-m) = (-mx + v_0)(-m)$
$a = m^2x - mv_0$
This is the equation of a straight line with a positive slope $(m^2)$ and a negative intercept $(-mv_0)$ on the acceleration axis.
Comparing this with the given options,graph $(A)$ represents a straight line with a negative intercept and a positive slope.
Therefore,option $(A)$ is correct.

(D) In a distance-time graph,the instantaneous velocity of a particle at any point is given by the slope of the tangent to the curve at that point.
Mathematically,$v = \frac{ds}{dt} = \tan(\theta)$,where $\theta$ is the angle that the tangent makes with the time axis.
The slope is maximum where the curve is steepest.
By observing the given graph,the curve is steepest at point $Q$. Therefore,the instantaneous velocity is maximum around point $Q$.

(D) By definition,the slope of a displacement-time graph represents the velocity of the particle.
$v = \tan \theta$
Given the angles $\theta_1 = 30^{\circ}$ and $\theta_2 = 45^{\circ}$.
The ratio of their velocities is given by:
$\frac{v_1}{v_2} = \frac{\tan \theta_1}{\tan \theta_2}$
Substituting the values:
$\frac{v_1}{v_2} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and $\tan 45^{\circ} = 1$,we get:
$\frac{v_1}{v_2} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$
Thus,the ratio is $1: \sqrt{3}$.

(A) The $v$-$t$ graph of the particle is shown in the figure.
$1$. For the first part (uniform acceleration):
Distance $2L = \frac{1}{2} \times V_m \times t_1 \implies t_1 = \frac{4L}{V_m}$
$2$. For the second part (constant velocity):
Distance $L = V_m \times t_2 \implies t_2 = \frac{L}{V_m}$
$3$. For the third part (uniform retardation):
Distance $3L = \frac{1}{2} \times V_m \times t_3 \implies t_3 = \frac{6L}{V_m}$
$4$. Average speed $\bar{V} = \frac{\text{Total distance}}{\text{Total time}}$
Total distance $= 2L + L + 3L = 6L$
Total time $= t_1 + t_2 + t_3 = \frac{4L}{V_m} + \frac{L}{V_m} + \frac{6L}{V_m} = \frac{11L}{V_m}$
$5$. Therefore,$\bar{V} = \frac{6L}{11L/V_m} = \frac{6}{11} V_m$
$\frac{\bar{V}}{V_m} = \frac{6}{11}$

(C) The velocity of a body projected vertically upwards with an initial velocity $v_{i}$ is given by the equation of motion:
$v_{f} = v_{i} - gt$
where $g$ is the acceleration due to gravity and $t$ is the time.
$1$. Initially,the velocity is positive $(v_{i})$ and decreases linearly with time as the body moves upwards.
$2$. At the maximum height,the velocity becomes zero $(v_{f} = 0)$ at time $t = \frac{v_{i}}{g}$.
$3$. After reaching the maximum height,the body starts moving downwards,so the velocity becomes negative and increases in magnitude linearly with time.
This behavior is represented by a straight line with a constant negative slope $(-g)$ that passes through the positive $v$-axis,crosses the $t$-axis at $t = \frac{v_{i}}{g}$,and continues into the negative $v$-region. Graph $C$ correctly depicts this linear decrease in velocity from a positive value to a negative value.

(C) We know that uniform velocity is independent of time. Therefore,in a velocity-time graph,the graph is represented as a straight line parallel to the time axis.
Since the slope of a line is given by $\text{slope} = \tan \theta$,where $\theta$ is the angle the line makes with the horizontal time axis.
From the graph,the line is parallel to the time axis,so $\theta = 0^{\circ}$.
Therefore,$\text{slope} = \tan 0^{\circ} = 0$.

(C) In the given graph,the $y$-axis represents time $(t)$ and the $x$-axis represents displacement $(s)$.
For a physical motion,time must always increase as the object moves.
In the given graph,as the object moves from $B$ to $C$ and then to $D$,the displacement $s$ decreases while time $t$ continues to increase.
However,at any given time $t$,an object can only have one position $s$.
Looking at the graph,for a single value of time $t$ (on the vertical axis),there are multiple values of displacement $s$ (on the horizontal axis) possible,which is physically impossible for a single object in motion.
Furthermore,the graph shows the object moving backward in displacement,which is possible,but the specific shape implies that for a single time instant,the object is at multiple positions,or more accurately,the slope $\frac{dt}{ds}$ is plotted instead of $\frac{ds}{dt}$.
Since time cannot decrease or be multi-valued for a single position,this graph is physically impossible.

(A) The graph is a circle with center $(2, 0)$ and radius $2 \ m$. Its equation is given by:
$(s-2)^2 + v^2 = 2^2$
$(s-2)^2 + v^2 = 4$
Differentiating both sides with respect to time $t$,we get:
$2(s-2) \frac{ds}{dt} + 2v \frac{dv}{dt} = 0$
Since $\frac{ds}{dt} = v$ and $\frac{dv}{dt} = a$,we substitute these into the equation:
$2(s-2)v + 2v \cdot a = 0$
Dividing by $2v$ (assuming $v \neq 0$):
$(s-2) + a = 0$
$a = -(s-2) = 2-s$
At the point $(s, v) = (2-\sqrt{2}, \sqrt{2})$:
$a = 2 - (2-\sqrt{2}) = \sqrt{2} \ ms^{-2}$

(A) The total distance covered is equal to the area under the $v-t$ graph.
From the graph,the maximum velocity $V_{max}$ is reached at time $t_1$.
We have $\tan \theta_1 = \alpha = \frac{V_{max}}{t_1} \implies t_1 = \frac{V_{max}}{\alpha}$.
Similarly,$\tan \theta_2 = \beta = \frac{V_{max}}{t_2} \implies t_2 = \frac{V_{max}}{\beta}$.
The total time is $t = t_1 + t_2 = V_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = V_{max} \left( \frac{\alpha + \beta}{\alpha \beta} \right)$.
Thus,$V_{max} = \frac{\alpha \beta t}{\alpha + \beta}$.
The total distance $S$ is the area of the triangle with base $t$ and height $V_{max}$:
$S = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times V_{max}$.
Substituting $V_{max}$,we get $S = \frac{1}{2} \times t \times \left( \frac{\alpha \beta t}{\alpha + \beta} \right) = \frac{1}{2} \left( \frac{\alpha \beta}{\alpha + \beta} \right) t^2$.

(B) The average velocity is defined as the total displacement divided by the total time interval: $v_{avg} = \frac{\Delta x}{\Delta t}$.
Displacement $\Delta x$ is the area under the velocity-time graph between $t=4 \,s$ and $t=6 \,s$.
At $t=6 \,s$, $v=15 \,ms^{-1}$. The graph is a straight line from $(0,0)$ to $(6,15)$, so the equation for velocity is $v(t) = \frac{15}{6}t = 2.5t$.
At $t=4 \,s$, $v(4) = 2.5 \times 4 = 10 \,ms^{-1}$.
The area under the graph between $t=4 \,s$ and $t=6 \,s$ is a trapezoid with parallel sides $v(4)=10 \,ms^{-1}$ and $v(6)=15 \,ms^{-1}$, and height $\Delta t = 6-4 = 2 \,s$.
Area = $\frac{1}{2} \times (v(4) + v(6)) \times \Delta t = \frac{1}{2} \times (10 + 15) \times 2 = 25 \,m$.
Average velocity $v_{avg} = \frac{25 \,m}{2 \,s} = 12.5 \,ms^{-1}$.
Thus, the correct option is $B$.

(B) The initial distance between the trains is $d_0 = 300 \, m$.
The distance covered by a train is equal to the area under its velocity-time graph.
For Train $I$, the area under the $v-t$ graph is $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \, s \times 40 \, m/s = 200 \, m$.
For Train $II$, the magnitude of the area under the $v-t$ graph is $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \, s \times 20 \, m/s = 80 \, m$.
Since the trains are moving in opposite directions towards each other, the total distance covered by both trains before stopping is $d_{total} = A_1 + A_2 = 200 \, m + 80 \, m = 280 \, m$.
The final separation between the trains when both have stopped is $d_{final} = d_0 - d_{total} = 300 \, m - 280 \, m = 20 \, m$.

(D) The acceleration $a$ is given by $a = v \frac{dv}{dx}$.
From the given graph,the velocity $v$ is a linear function of displacement $x$.
For the interval $0 \le x \le 100$,the slope is positive. Let $v = kx$ (where $k > 0$). Then,$a = v \frac{dv}{dx} = (kx)(k) = k^2 x$. This shows that $a$ increases linearly with $x$ in this interval.
For the interval $100 \le x \le 200$,the slope is negative. Let $v = -k(x - 200) = -kx + 200k$. Then,$a = v \frac{dv}{dx} = (-kx + 200k)(-k) = k^2 x - 200k^2$. This shows that $a$ increases linearly with $x$ in this interval,but with a negative intercept.
Comparing this with the given options,the graph that shows a positive linear increase in acceleration for the first part and a positive linear increase with a negative value for the second part is option $D$.

(C) Velocity is defined as the rate of change of displacement.
$v_y = \frac{dy}{dt}$
$= \frac{d(P t^2 - Q t^3)}{dt}$
$= 2Pt - 3Qt^2$
At maximum height,the vertical velocity of the particle becomes zero.
$2Pt - 3Qt^2 = 0$
$t(2P - 3Qt) = 0$
Since $t=0$ is the start of the motion,the time at maximum height is $t = \frac{2P}{3Q}$.
Now,substituting this time into the displacement equation:
$y_{max} = P(\frac{2P}{3Q})^2 - Q(\frac{2P}{3Q})^3$
$= P(\frac{4P^2}{9Q^2}) - Q(\frac{8P^3}{27Q^3})$
$= \frac{4P^3}{9Q^2} - \frac{8P^3}{27Q^2}$
$= \frac{12P^3 - 8P^3}{27Q^2}$
$= \frac{4P^3}{27Q^2}$

(D) The velocity $v$ of a particle is given by the slope of the displacement-time graph,which is $v = \tan(\theta)$,where $\theta$ is the angle the graph makes with the time axis.
For the first particle,$\theta_1 = 30^{\circ}$,so its velocity $v_1 = \tan(30^{\circ}) = \frac{1}{\sqrt{3}}$.
For the second particle,$\theta_2 = 45^{\circ}$,so its velocity $v_2 = \tan(45^{\circ}) = 1$.
The ratio of their velocities is $\frac{v_1}{v_2} = \frac{\tan(30^{\circ})}{\tan(45^{\circ})} = \frac{1/\sqrt{3}}{1} = 1 : \sqrt{3}$.

(B) As we know that, distance is equal to the area under the velocity-time graph. Since average speed is defined as the total distance traveled divided by the total time taken, we must consider the magnitude of the area for both positive and negative velocity intervals.
Total distance traveled in $80 \, s$ is the sum of the absolute areas of the triangles formed:
$Distance = |Area_{OAB}| + |Area_{BCD}|$
Area of triangle $OAB$ (from $t=0$ to $t=40 \, s$):
$Area_{OAB} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \, s \times 10 \, m/s = 200 \, m$
Area of triangle $BCD$ (from $t=40$ to $t=80 \, s$):
$Area_{BCD} = \frac{1}{2} \times \text{base} \times |\text{height}| = \frac{1}{2} \times 40 \, s \times |-10 \, m/s| = 200 \, m$
Total distance $= 200 \, m + 200 \, m = 400 \, m$
Total time taken $= 80 \, s$
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{400 \, m}{80 \, s} = 5 \, m/s$

(B) The distance-time graph for the motion of a particle from $A$ to $B$ is shown in the diagram.
From the graph,the slope of the distance-time graph for the segments $A$ to $C$ and $D$ to $B$ is positive,which means the speed is non-zero in these intervals.
Average speed is defined as the total distance covered divided by the total time taken. Since the total distance covered from $A$ to $B$ is non-zero,the average speed is always non-zero.
Instantaneous speed is the slope of the distance-time graph at any given instant. In the segment $C$ to $D$,the graph is a horizontal line,meaning the distance does not change with time. Therefore,the slope is zero,which implies the instantaneous speed is zero during this interval.
Thus,the average speed is always non-zero,but the instantaneous speed can be zero. Hence,option $B$ is correct.

(C) The position of the particle is given by the function $y(t) = 10 t e^{-2 t}$.
To find when the particle stops momentarily,we calculate its velocity $v$ by differentiating $y(t)$ with respect to time $t$:
$v = \frac{dy}{dt} = \frac{d}{dt} (10 t e^{-2 t})$.
Using the product rule,$v = 10 [t \cdot (-2 e^{-2 t}) + e^{-2 t} \cdot 1] = 10 e^{-2 t} (1 - 2 t)$.
The particle stops momentarily when $v = 0$,which implies $1 - 2 t = 0$,so $t = \frac{1}{2} \ s$.
Now,substitute $t = \frac{1}{2} \ s$ into the position equation to find the distance from the origin:
$y(\frac{1}{2}) = 10 \times (\frac{1}{2}) \times e^{-2 \times (\frac{1}{2})} = 5 \times e^{-1} = \frac{5}{e} \ m$.

(B) The velocity of the bird is given by $v(t) = t - 2$.
Since distance is the integral of the magnitude of velocity, we calculate:
$\text{Distance} = \int_{0}^{4} |v(t)| \text{ dt} = \int_{0}^{4} |t - 2| \text{ dt}$.
We split the integral at $t = 2 \text{ s}$ where the velocity changes sign:
$\text{Distance} = \int_{0}^{2} -(t - 2) \text{ dt} + \int_{2}^{4} (t - 2) \text{ dt}$.
Evaluating the first part:
$\int_{0}^{2} (2 - t) \text{ dt} = [2t - \frac{t^2}{2}]_{0}^{2} = (4 - 2) - 0 = 2 \text{ m}$.
Evaluating the second part:
$\int_{2}^{4} (t - 2) \text{ dt} = [\frac{t^2}{2} - 2t]_{2}^{4} = (8 - 8) - (2 - 4) = 0 - (-2) = 2 \text{ m}$.
Total distance $= 2 \text{ m} + 2 \text{ m} = 4 \text{ m}$.

(A) The velocity-time graph is a trapezoid. Let the car accelerate for time $t_1$ and decelerate for time $t_1$. The time for which it moves with uniform velocity is $(25 - 2t_1) \ s$.
Maximum velocity $v_{max} = a \times t_1 = 5t_1$.
Average velocity $v_{avg} = 72 \ km \ hr^{-1} = 72 \times \frac{5}{18} \ m \ s^{-1} = 20 \ m \ s^{-1}$.
Total displacement $S = v_{avg} \times t_{total} = 20 \times 25 = 500 \ m$.
The area under the velocity-time graph is the displacement:
$S = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$500 = \frac{1}{2} \times [ (25 - 2t_1) + 25 ] \times 5t_1$
$1000 = (50 - 2t_1) \times 5t_1$
$1000 = 250t_1 - 10t_1^2$
$10t_1^2 - 250t_1 + 1000 = 0$
$t_1^2 - 25t_1 + 100 = 0$
$(t_1 - 20)(t_1 - 5) = 0$
Since $t_1$ must be less than $12.5 \ s$ (as $2t_1 < 25$),we have $t_1 = 5 \ s$.
The time for which the car moved with uniform velocity is $25 - 2t_1 = 25 - 2(5) = 15 \ s$.

(B) The equation of the $v$ vs $x$ graph is a straight line with a negative slope and a positive intercept on the $v$-axis:
$v = C_1 - C_2 x$,where $C_1$ and $C_2$ are positive constants.
The acceleration $a$ is given by the relation:
$a = v \frac{dv}{dx}$
Substituting the expression for $v$ and its derivative $\frac{dv}{dx} = -C_2$:
$a = (C_1 - C_2 x) \times (-C_2)$
$a = C_2^2 x - C_1 C_2$
This equation is of the form $a = mx + c$,where the slope $m = C_2^2$ (positive) and the intercept $c = -C_1 C_2$ (negative).
Thus,the graph of $a$ versus $x$ is a straight line with a positive slope and a negative intercept on the $a$-axis,which corresponds to option $B$.

(C) Distance is the total path length covered,which is the sum of the areas under the $|v|$ vs $t$ graph.
For the interval $0$ to $20 \text{ s}$,the area of the triangle is $A_1 = 1/2 \times 20 \times 5 = 50 \text{ m}$.
For the interval $20$ to $40 \text{ s}$,the magnitude of velocity is $5 \text{ m/s}$,so the area is $A_2 = 1/2 \times 20 \times 5 = 50 \text{ m}$.
Total distance $= A_1 + A_2 = 50 + 50 = 100 \text{ m}$.
Displacement is the net change in position,which is the algebraic sum of the areas under the $v$ vs $t$ graph.
Displacement $= A_1 - A_2 = 50 - 50 = 0 \text{ m}$.
Average velocity $= \text{Displacement} / \text{Total time} = 0 / 40 = 0 \text{ m/s}$.

(C) When a ball is thrown vertically upward with an initial velocity $u$,its velocity at any time $t$ is given by the equation of motion: $v = u - gt$,where $g$ is the acceleration due to gravity.
$1$. Initially,the velocity is positive $(v = u)$.
$2$. As the ball rises,the velocity decreases linearly with time because of the constant downward acceleration $g$.
$3$. At the maximum height,the velocity becomes zero.
$4$. After reaching the maximum height,the ball starts falling downward. During this phase,the velocity becomes negative (as it is directed opposite to the initial upward direction) and its magnitude increases linearly with time.
This linear relationship $v = u - gt$ represents a straight line with a negative slope $(-g)$. Plot $(C)$ correctly depicts this variation,starting from a positive initial velocity,crossing the time axis (where $v = 0$),and continuing into the negative velocity region.