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

(D) The velocity of a particle in a displacement-time graph is given by the slope of the line,which is $\tan \theta$,where $\theta$ is the angle the line makes with the time axis.
For particle $A$,the angle is $\theta_A = 30^\circ$,so $V_A = \tan 30^\circ = 1/\sqrt{3}$.
For particle $B$,the angle is $\theta_B = 60^\circ$,so $V_B = \tan 60^\circ = \sqrt{3}$.
The ratio of velocities is $\frac{V_A}{V_B} = \frac{\tan 30^\circ}{\tan 60^\circ} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.
Therefore,the ratio $V_A:V_B$ is $1:3$.

(D) The tangent to a curve at any instant on an $x-t$ (position-time) graph gives the instantaneous velocity of the body.
$A$ speedometer in a vehicle reads the instantaneous velocity of the body at every instant.
Thus,the instantaneous velocity of the body can be measured both graphically and by a speedometer.

(B) The displacement of the particle is given by $s = 6t^2 - t^3$.
Velocity $v$ is the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$.
$v = \frac{d}{dt}(6t^2 - t^3) = 12t - 3t^2$.
To find the time when the velocity is zero,we set $v = 0$:
$12t - 3t^2 = 0$.
$3t(4 - t) = 0$.
This gives $t = 0 \ s$ and $t = 4 \ s$.
The particle starts from rest at $t = 0 \ s$,so it will attain zero velocity again at $t = 4 \ s$.

(B) The position of the particle is given by the equation $x = a + bt^2$.
To find the instantaneous velocity $v$,we differentiate the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2)$
Since $a$ is a constant,its derivative is $0$. Thus,$v = 2bt$.
Given $b = 3 \, cm/s^2$ and $t = 3 \, s$,we substitute these values into the velocity equation:
$v = 2 \times 3 \times 3 = 18 \, cm/s$.
Therefore,the instantaneous velocity at $t = 3 \, s$ is $18 \, cm/s$.

(D) Given the relation: $3t = \sqrt{3x} + 6$.
Rearranging for $\sqrt{3x}$: $\sqrt{3x} = 3t - 6$.
Squaring both sides: $3x = (3t - 6)^2 = 9t^2 - 36t + 36$.
Dividing by $3$: $x = 3t^2 - 12t + 12$.
Velocity $v$ is the derivative of displacement $x$ with respect to time $t$: $v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 12t + 12) = 6t - 12$.
For velocity to be zero: $6t - 12 = 0 \Rightarrow t = 2 \ s$.
Substituting $t = 2 \ s$ into the displacement equation: $x = 3(2)^2 - 12(2) + 12 = 12 - 24 + 12 = 0 \ m$.

(A) The velocity $v$ of the particle is given by the derivative of the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(2 - 5t + 6t^2)$
$v = 0 - 5 + 12t$
$v = (12t - 5) \ m/s$
To find the initial velocity,we substitute $t = 0$ into the velocity equation:
$v(0) = 12(0) - 5 = -5 \ m/s$
Thus,the initial velocity of the particle is $-5 \ m/s$.

(D) Given displacement $x = a e^{-\alpha t} + b e^{\beta t}$.
Velocity $v$ is the rate of change of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t})$
$v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$.
To determine how velocity changes with time,we find the acceleration $a_{acc}$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(-a\alpha e^{-\alpha t} + b\beta e^{\beta t})$
$a_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$.
Since $a, b, \alpha, \beta$ are positive constants and the exponential functions $e^{-\alpha t}$ and $e^{\beta t}$ are always positive for any time $t$,the acceleration $a_{acc}$ is always positive.
Because the acceleration is positive,the velocity $v$ will go on increasing with time.

(B) The distance travelled by a particle is equal to the area under the velocity-time $(v-t)$ graph.
From the given graph,the total distance is the sum of the areas of the four geometric shapes $(A_1, A_2, A_3, A_4)$:
$1$. Area $A_1$ (triangle from $t=0$ to $t=1$): $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 20 = 10 \; m$.
$2$. Area $A_2$ (rectangle from $t=1$ to $t=2$): $\text{length} \times \text{breadth} = 1 \times 20 = 20 \; m$.
$3$. Area $A_3$ (trapezium from $t=2$ to $t=3$): $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (20 + 10) \times 1 = 15 \; m$.
$4$. Area $A_4$ (rectangle from $t=3$ to $t=4$): $\text{length} \times \text{breadth} = 1 \times 10 = 10 \; m$.
Total distance = $A_1 + A_2 + A_3 + A_4 = 10 + 20 + 15 + 10 = 55 \; m$.

(A) The slope of the displacement-time graph represents the velocity of the particle $(v = dx/dt)$.
As time increases,the slope of the curve continuously decreases,which indicates that the velocity of the particle is decreasing.
This means the motion is retarded (decelerated).
Finally,the slope of the curve becomes zero at the horizontal asymptote,which implies that the velocity becomes zero and the particle stops.
Therefore,option $(A)$ is correct.

(D) When a ball is thrown vertically upwards with an initial velocity $u$,it moves under the influence of constant gravitational acceleration $g$ acting downwards.
According to the equation of motion,$v = u - gt$.
$1$. During the upward journey,the velocity is positive and decreases linearly with time until it becomes zero at the maximum height.
$2$. At the maximum height,the velocity is zero.
$3$. During the downward journey,the velocity becomes negative (as it is directed downwards) and its magnitude increases linearly with time.
$4$. The slope of the velocity-time graph is $-g$,which is constant throughout the motion.
Therefore,the graph is a straight line with a negative slope passing through the positive velocity axis,crossing the time axis,and continuing into the negative velocity region. This corresponds to the graph shown in option $D$.

(B) The acceleration of a particle is given by the second derivative of displacement with respect to time,$a = \frac{d^2x}{dt^2}$,which corresponds to the concavity of the $x-t$ graph.
$1$. Region $OA$: The graph is concave downwards (bending towards the time axis). The slope (velocity) is decreasing,so the acceleration is negative $(-)$.
$2$. Region $AB$: The graph is a straight line parallel to the time axis. The slope (velocity) is constant and zero. Therefore,the acceleration is zero $(0)$.
$3$. Region $BC$: The graph is concave upwards (bending towards the displacement axis). The slope (velocity) is increasing,so the acceleration is positive $(+)$.
$4$. Region $CD$: The graph is a straight line with a constant non-zero slope. Since the velocity is constant,the acceleration is zero $(0)$.
Thus,the signs of acceleration for $OA, AB, BC, CD$ are $-, 0, +, 0$ respectively. The correct option is $(b)$.

(D) Maximum acceleration corresponds to the steepest slope of the $v - t$ graph,which represents the maximum rate of change of velocity.
$1$. For the interval $t = 0$ to $t = 20 \, sec$: $a = \frac{20 - 0}{20 - 0} = 1 \, cm/sec^2$.
$2$. For the interval $t = 20$ to $t = 30 \, sec$: $a = 0 \, cm/sec^2$ (constant velocity).
$3$. For the interval $t = 30$ to $t = 40 \, sec$: $a = \frac{80 - 20}{40 - 30} = \frac{60}{10} = 6 \, cm/sec^2$.
$4$. For the interval $t = 40$ to $t = 80 \, sec$: $a = \frac{0 - 80}{80 - 40} = \frac{-80}{40} = -2 \, cm/sec^2$.
Comparing the magnitudes of acceleration in all intervals,the maximum acceleration is $6 \, cm/sec^2$. Thus,the correct option is $D$.

(D) In an $x - t$ (displacement-time) graph,the slope represents the velocity of the body.
For $t < t_1$,the graph is a straight line with a constant positive slope,which indicates that the body is moving with a constant velocity.
For $t > t_1$,the graph is a horizontal line,meaning the slope is zero. This indicates that the displacement of the body is not changing,so the velocity is zero (the body is at rest).
Therefore,the body travels with a constant speed up to time $t_1$ and then stops.

(C) The total height (displacement) reached by the lift is equal to the area under the velocity-time graph.
The graph is a trapezium with parallel sides of lengths $10 - 2 = 8$ and $12 - 0 = 12$, and a height (velocity) of $3.6$.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
Area $= \frac{1}{2} \times (8 + 12) \times 3.6$
Area $= \frac{1}{2} \times 20 \times 3.6 = 10 \times 3.6 = 36\,m$.

(A) Displacement is the algebraic sum of the areas under the velocity-time graph.
Displacement $= A_1 + (-A_2) + A_3$
$= (2 \, s \times 4 \, m/s) + (2 \, s \times -2 \, m/s) + (2 \, s \times 2 \, m/s)$
$= 8 \, m - 4 \, m + 4 \, m = 8 \, m$
Distance is the sum of the magnitudes of the areas under the velocity-time graph.
Distance $= |A_1| + |A_2| + |A_3|$
$= |8 \, m| + |-4 \, m| + |4 \, m|$
$= 8 \, m + 4 \, m + 4 \, m = 16 \, m$
Thus,the displacement is $8 \, m$ and the distance is $16 \, m$.

(B) The interval with non-zero acceleration and retardation is from $t = 20 \, s$ to $t = 40 \, s$.
Displacement is equal to the area under the $v-t$ graph.
For the interval $t = 20 \, s$ to $t = 40 \, s$,the area consists of a rectangle of height $1 \, m/s$ and width $20 \, s$,plus a triangle of base $20 \, s$ and height $(4 - 1) = 3 \, m/s$.
Area $= (20 \times 1) + (\frac{1}{2} \times 20 \times 3)$
Area $= 20 + 30 = 50 \, m$.

(C) The displacement-time equation for uniform acceleration is given by $x = x_0 + u t + \frac{1}{2} a t^2$.
For a particle starting from rest $(u=0)$,this simplifies to $x = x_0 + \frac{1}{2} a t^2$,which represents a parabolic curve.
In figure $(i)$,the slope of the displacement-time graph (which represents velocity) increases with time,indicating that the particle is undergoing uniformly accelerated motion.
In figure $(ii)$,the slope of the displacement-time graph decreases with time,indicating that the particle is undergoing uniformly retarded (decelerated) motion.
Therefore,particle $(i)$ is having a uniformly accelerated motion while particle $(ii)$ is having a uniformly retarded motion.

(B) The distance covered by an object in a velocity-time graph is equal to the area under the graph.
$1$. Total distance covered in $7 \text{ s}$ (Area of the trapezium):
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$\text{Area} = \frac{1}{2} \times (7 - 1 + 5 - 3) \times 10 = \frac{1}{2} \times (6 + 2) \times 10 = \frac{1}{2} \times 8 \times 10 = 40 \text{ m}$.
$2$. Distance covered in the last $2 \text{ s}$ (from $t = 5 \text{ s}$ to $t = 7 \text{ s}$):
This is the area of the triangle formed between $t = 5$ and $t = 7$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (7 - 5) \times 10 = \frac{1}{2} \times 2 \times 10 = 10 \text{ m}$.
$3$. Required fraction:
$\text{Fraction} = \frac{\text{Distance in last } 2 \text{ s}}{\text{Total distance}} = \frac{10}{40} = 0.25$.

(A) The distance travelled by a body in a velocity-time graph is equal to the area under the velocity-time curve.
Here,the shape under the graph is a trapezium with parallel sides of length $10 \; s$ (from $10 \; s$ to $20 \; s$) and $30 \; s$ (from $0 \; s$ to $30 \; s$),and a height of $10 \; m/s$.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Distance $= \frac{1}{2} \times (10 + 30) \times 10$
Distance $= \frac{1}{2} \times 40 \times 10 = 200 \; m$.

(D) The instantaneous velocity of a particle is given by the slope of the displacement-time graph at that point.
Mathematically,$v = \frac{ds}{dt}$.
- At point $C$,the slope is positive as the displacement is increasing.
- At point $D$,the slope is zero because it is at the peak of the curve.
- At point $E$,the slope is negative because the displacement is decreasing with time.
- At point $F$,the slope is positive as the displacement is increasing again.
Therefore,the instantaneous velocity is negative at point $E$.

(A) Uniform motion is defined as motion where an object covers equal distances in equal intervals of time. In a position-time $(s-t)$ graph,this corresponds to a straight line with a constant slope,as the velocity $(v = \frac{ds}{dt})$ remains constant. Graph $(A)$ shows a linear relationship between position and time,indicating a constant velocity,which represents uniform motion.

(A) The velocity $v$ of a body is given by the slope of the displacement-time graph,i.e.,$v = \frac{ds}{dt}$.
$1$. Initially,the slope of the displacement-time graph is positive and decreasing,which means the velocity is positive and decreasing.
$2$. At the peak of the graph,the slope is zero,so the velocity is zero.
$3$. After the peak,the slope becomes negative,which means the velocity becomes negative.
$4$. Among the given options,the graph that shows velocity starting from a positive value,decreasing to zero,and then becoming negative is represented by option $A$.

(B) The acceleration is given by the slope of the speed-time graph.
For part $AB$: Slope $= \frac{20 - 0}{0.75 - 0} = \frac{20}{0.75} \approx 26.67 \; km\, h^{-2}$.
For part $BC$: Slope $= 0$ (constant speed).
For part $CD$: Slope $= \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160 \; km\, h^{-2}$.
For part $DE$: Slope $= \frac{0 - 60}{2.00 - 1.25} = \frac{-60}{0.75} = -80 \; km\, h^{-2}$ (deceleration).
Comparing these values,the maximum acceleration is $160 \; km\, h^{-2}$,which corresponds to the segment $CD$.

(D) The correct option is $D$.
By definition,acceleration $a$ is the rate of change of velocity with respect to time,given by the formula $a = \frac{dv}{dt}$.
Rearranging this equation,we get $dv = a \cdot dt$.
To find the total change in velocity over a time interval from $t_1$ to $t_2$,we integrate both sides: $\Delta v = \int_{t_1}^{t_2} a \cdot dt$.
Geometrically,the integral $\int a \cdot dt$ represents the area under the acceleration-time graph.
Therefore,the area under the acceleration-time graph gives the change in velocity of the object.

(A) For uniform acceleration,the equation of motion is given by $s = ut + \frac{1}{2}at^2$.
If the initial velocity $u = 0$,then $s = \frac{1}{2}at^2$.
This equation represents a parabola where $s \propto t^2$.
Graph $(A)$ shows a parabolic curve,which is characteristic of motion with constant (uniform) acceleration.

(B) For a body in motion,at any single instant of time $t$,there can be only one unique value of velocity $v$.
In graphs $A$,$C$,and $D$,for a single value of time $t$,the graph shows multiple values of velocity $v$,which is physically impossible.
Graph $B$ represents a realistic situation where the velocity varies continuously with time,having a unique value for every instant $t$.

(A) Uniform motion is defined as motion with a constant velocity,meaning the acceleration is zero $(a = 0)$.
In a velocity-time $(v-t)$ graph,the slope represents acceleration.
For graph $(A)$,the velocity remains constant over time,resulting in a horizontal line parallel to the time axis. The slope of this line is zero,which implies $a = 0$. Therefore,graph $(A)$ represents uniform motion.
Graphs $(B)$,$(C)$,and $(D)$ represent non-uniform motion because their slopes are non-zero or changing,indicating acceleration.

(D) From the acceleration-time graph,we can observe three distinct phases of motion:
$1$. In the first phase,the acceleration $a$ is constant and positive. Since $v = \int a \, dt$,the velocity increases linearly with time,represented by a straight line with a positive slope on a velocity-time graph.
$2$. In the second phase,the acceleration $a = 0$. This means the velocity remains constant,represented by a horizontal line on a velocity-time graph.
$3$. In the third phase,the acceleration $a$ is again constant and positive. This means the velocity increases linearly with time again,represented by a straight line with a positive slope.
Comparing this with the given options,the graph that shows a linear increase in velocity,followed by a constant velocity,and then another linear increase is option $D$.

(B) The given graph is a straight line with a positive intercept $v_0$ and a negative slope. The equation of the line is given by:
$v = -mx + v_0$ ... $(i)$
where $m = \tan \theta = \frac{v_0}{x_0}$ is the magnitude of the slope.
We know that acceleration $a = v \frac{dv}{dx}$.
From equation $(i)$,differentiating with respect to $x$ gives:
$\frac{dv}{dx} = -m$
Substituting this into the acceleration formula:
$a = (-mx + v_0)(-m)$
$a = m^2x - mv_0$
This is the equation of a straight line with a positive slope $(m^2)$ and a negative intercept $(-mv_0)$ on the $a$-axis.
Comparing this with the given options,Graph $B$ represents a straight line with a negative intercept on the $a$-axis and a positive slope. Therefore,Graph $B$ is correct.

(D) From the given $a-t$ graph,it is clear that the acceleration is increasing at a constant rate.
$\therefore \frac{da}{dt} = k$ (constant) $\implies a = kt$ (by integration).
Since $a = \frac{dv}{dt}$,we have $\frac{dv}{dt} = kt \implies dv = kt \, dt$.
Integrating both sides: $\int dv = \int kt \, dt \implies v = \frac{1}{2}kt^2$.
This shows that $v$ is dependent on time parabolically,where the graph is a parabola opening upwards.
After the acceleration reaches its maximum value,it suddenly becomes zero. When acceleration is zero,the velocity of the body becomes constant.
Therefore,the velocity-time graph will show a parabolic increase followed by a horizontal line representing constant velocity. This corresponds to graph $(d)$.

(C) The slope of a displacement-time graph gives the velocity. However,the slope is defined as $\tan \theta$,where $\theta$ is the angle made by the line with the time axis.
In the given graph,the angle $30^{\circ}$ is made with the displacement axis.
The angle with the time axis is $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Therefore,the velocity $v = \tan(60^{\circ}) = \sqrt{3} \, m/s$.

(A) The average velocity is defined as the total displacement divided by the total time interval.
In a $v-t$ graph,the displacement is equal to the area under the curve.
For the interval $t = 0$ to $t = 5$ seconds,the area is a triangle with base $5$ and height $5$,so displacement $s_1 = \frac{1}{2} \times 5 \times 5 = 12.5 \ m$.
For the interval $t = 5$ to $t = 10$ seconds,the area is a triangle with base $5$ and height $-5$,so displacement $s_2 = \frac{1}{2} \times 5 \times (-5) = -12.5 \ m$.
Total displacement $s = s_1 + s_2 = 12.5 + (-12.5) = 0 \ m$.
Average velocity = $\frac{\text{Total displacement}}{\text{Total time}} = \frac{0}{10} = 0 \ m/s$.

(D) The distance traveled by the lift is equal to the area under the velocity-time graph.
The graph is a trapezium with parallel sides of lengths $12\, s$ and $(10 - 2) = 8\, s$,and a height of $3.6\, m/s$.
Alternatively,the area can be calculated as the sum of the areas of two triangles and one rectangle:
Area $= (\text{Area of first triangle}) + (\text{Area of rectangle}) + (\text{Area of second triangle})$
Area $= (\frac{1}{2} \times 2 \times 3.6) + (8 \times 3.6) + (\frac{1}{2} \times 2 \times 3.6)$
Area $= 3.6 + 28.8 + 3.6 = 36\, m$.
Thus,the total height reached by the lift is $36\, m$.

(D) According to Newton's first law of motion,if the net force acting on a particle is zero,it moves with a constant velocity.
In a displacement-time $(x-t)$ graph,the velocity is represented by the slope of the graph.
If the slope is constant,the velocity is constant,which implies that the acceleration is zero and the net force acting on the particle is zero.
In the given graph,the segments $AB$ and $CD$ are straight lines,meaning the slope (velocity) is constant in these regions.
Therefore,the force acting on the particle is zero in regions $AB$ and $CD$.

(C) When a batsman hits a ball,it moves upward with an initial vertical velocity. Under the influence of gravity,its vertical velocity decreases linearly with time until it reaches the highest point of its trajectory,where the vertical velocity becomes zero. After this point,the ball starts moving downward,and its vertical velocity increases in the downward direction (which is represented as negative velocity). The acceleration due to gravity $g$ is constant and acts downwards throughout the motion,meaning the slope of the velocity-time graph $(dv/dt = -g)$ is constant and negative. This linear decrease in velocity from a positive value to a negative value is correctly represented by graph $C$.

(D) Given equation: $t = \sqrt{x} + 3$
Rearranging for $x$: $\sqrt{x} = t - 3$
Squaring both sides: $x = (t - 3)^2$
Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t - 3)^2 = 2(t - 3)$
When velocity is zero: $2(t - 3) = 0 \implies t = 3 \text{ s}$
Position at $t = 3 \text{ s}$: $x = (3 - 3)^2 = 0 \text{ m}$
Initial position at $t = 0 \text{ s}$: $x_0 = (0 - 3)^2 = 9 \text{ m}$
Displacement $\Delta x = x - x_0 = 0 - 9 = -9 \text{ m}$
The magnitude of displacement is $9 \text{ m}$.

(B) The position of the particle is given by $x = a + bt^2$.
The velocity $v$ is the time derivative of the position $x$:
$v = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2)$
Since $a$ is a constant,its derivative is $0$:
$v = 0 + 2bt = 2bt$
Given $b = 3 \, cm/s^2$ and $t = 3 \, s$:
$v = 2 \times 3 \times 3$
$v = 18 \, cm/s$.

(D) The velocity is given by the slope of the displacement-time graph.
For the first $1 \ s$ ($t=0$ to $t=1$):
$v_1 = \frac{\Delta x}{\Delta t} = \frac{30 - 0}{1 - 0} = 30 \ m/s$.
For the next $2 \ s$ ($t=1$ to $t=3$):
$v_2 = \frac{\Delta x}{\Delta t} = \frac{0 - 30}{3 - 1} = \frac{-30}{2} = -15 \ m/s$.
The magnitude of velocity in the next $2 \ s$ is $|v_2| = 15 \ m/s$.
The ratio of the velocity in the first $1 \ s$ to the magnitude of velocity in the next $2 \ s$ is $\frac{30}{15} = \frac{2}{1}$,which is $2:1$.

(B) The acceleration is given by the slope of the speed-time graph.
For segment $AB$: $a_1 = \frac{20 - 0}{0.75 - 0} = \frac{20}{0.75} \approx 26.67 \ km \ h^{-2}$.
For segment $BC$: $a_2 = 0 \ km \ h^{-2}$ (constant speed).
For segment $CD$: $a_3 = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160 \ km \ h^{-2}$.
For segment $DE$: $a_4 = \frac{0 - 60}{2.00 - 1.25} = \frac{-60}{0.75} = -80 \ km \ h^{-2}$.
Comparing these values,the maximum acceleration is $160 \ km \ h^{-2}$.

(A) The velocity $v$ is given by the slope of the displacement-time $(s-t)$ graph,i.e.,$v = \frac{ds}{dt}$.
In the given $s-t$ graph,the curve is a parabola opening downwards,which can be represented by the equation $s = at - bt^2$,where $a$ and $b$ are positive constants.
Differentiating this with respect to time $t$,we get the velocity $v = \frac{ds}{dt} = a - 2bt$.
This equation $v = a - 2bt$ represents a straight line with a negative slope $(-2b)$ and a positive intercept $(a)$ on the velocity axis.
Thus,the velocity-time graph is a straight line with a negative slope,which matches the graph shown in option $A$.

(A) The distance covered by an object is equal to the area under the velocity-time $(v-t)$ graph.
The graph is a trapezium with parallel sides of lengths $30 \ s$ and $(20 - 10) = 10 \ s$,and a height of $10 \ m/s$.
Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Area = $\frac{1}{2} \times (30 + 10) \times 10$
Area = $\frac{1}{2} \times 40 \times 10 = 200 \ m$.
Therefore,the distance covered by the lift is $200 \ m$.

(B) The distance covered is equal to the area under the velocity-time graph.
Total distance covered in $7\,s$ is the area of the trapezium $ABCD$:
Area $= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Area $= \frac{1}{2} \times (6 + 2) \times 10 = \frac{1}{2} \times 8 \times 10 = 40\,m$.
Distance covered in the last $2\,s$ (from $t = 5\,s$ to $t = 7\,s$) is the area of the triangle $CQD$:
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times (7 - 5) \times 10 = \frac{1}{2} \times 2 \times 10 = 10\,m$.
The required fraction is $\frac{10}{40} = 0.25$.

(B) The given acceleration-time $(a-t)$ graph shows that acceleration increases linearly with time $(a \propto t)$ for a certain interval and then becomes zero (or drops to zero).
$1$. For the interval where acceleration increases linearly: $a = kt$ (where $k$ is a constant).
Since $a = \frac{dv}{dt}$,we have $\frac{dv}{dt} = kt$.
Integrating both sides with respect to time: $\int dv = \int kt \, dt$.
This gives $v = \frac{1}{2} kt^2 + C$. Since the graph starts from the origin,$C = 0$,so $v = \frac{1}{2} kt^2$.
This represents a parabolic curve opening upwards.
$2$. When the acceleration becomes zero,the velocity becomes constant (as $\frac{dv}{dt} = 0$).
Therefore,the velocity-time graph should show a parabolic increase followed by a horizontal line representing constant velocity. This matches the graph in option $B$.

(D) In a velocity-time graph,the velocity $v$ is represented on the $y$-axis and time $t$ on the $x$-axis.
For a graph to be physically possible,for every instant of time $t$,there must be a unique value of velocity $v$.
In graph $D$,for a single instant of time $t$,there are two different values of velocity $v$,which is physically impossible.
Therefore,graph $D$ is not possible.

(A) When a steel ball is dropped from a height,it accelerates downwards due to gravity,so its velocity increases in the negative direction (downward). This is represented by a straight line with a negative slope starting from the origin.
When it hits the marble floor,it undergoes an elastic collision,causing an instantaneous change in the direction of velocity from negative to positive. This is represented by a vertical jump in the graph.
After the bounce,it moves upwards with a positive velocity that decreases due to gravity,represented by a straight line with a negative slope in the positive velocity region.
As the ball continues to bounce,the height of each subsequent bounce decreases due to energy loss,resulting in smaller velocity changes and shorter time intervals between bounces. This pattern of linear segments with negative slopes and vertical jumps is correctly depicted in option $A$.

(A) The acceleration is given by the slope of the velocity-time graph.
For segment $OA$: Acceleration $a_{OA} = \frac{v_A - v_O}{t_A - t_O} = \frac{10 - 0}{10 - 0} = 1 \ m/s^2$.
For segment $AB$: Since velocity is constant,acceleration $a_{AB} = 0 \ m/s^2$.
For segment $BC$: Acceleration $a_{BC} = \frac{v_C - v_B}{t_C - t_B} = \frac{0 - 10}{40 - 20} = \frac{-10}{20} = -0.5 \ m/s^2$.
Thus,the accelerations are $1, 0, -0.5$.

(A) Given the position equation: $x = 9t^2 - t^3$.
Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2$.
To find the time at which speed is maximum,we differentiate velocity with respect to time and set it to zero: $\frac{dv}{dt} = 18 - 6t = 0$.
Solving for $t$,we get $t = 3 \ s$.
Now,substitute $t = 3 \ s$ into the position equation to find the position at that instant: $x = 9(3)^2 - (3)^3 = 9(9) - 27 = 81 - 27 = 54 \ m$.

(B) The velocity of the particle is given as $v(x) = \beta x^{-2n}$.
To find the acceleration $a$,we use the relation $a = v \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx} (\beta x^{-2n}) = \beta (-2n) x^{-2n-1} = -2n \beta x^{-2n-1}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (\beta x^{-2n}) \times (-2n \beta x^{-2n-1})$.
$a = -2n \beta^2 x^{-2n + (-2n) - 1}$.
$a = -2n \beta^2 x^{-4n-1}$.

(B) The position of car $P$ at any time $t$ is given by $x_P(t) = at + bt^2$.
The velocity of car $P$ is $v_P(t) = \frac{dx_P(t)}{dt} = a + 2bt$ ... $(i)$.
Similarly,for car $Q$,the position is $x_Q(t) = ft - t^2$.
The velocity of car $Q$ is $v_Q(t) = \frac{dx_Q(t)}{dt} = f - 2t$ ... $(ii)$.
Given that the cars have the same velocity,$v_P(t) = v_Q(t)$.
Therefore,$a + 2bt = f - 2t$.
Rearranging the terms to solve for $t$: $2bt + 2t = f - a$.
$2t(b + 1) = f - a$.
Thus,$t = \frac{f - a}{2(1 + b)}$.