Acceleration and its graph Questions in English

(C) The velocity of the body is given by $v = 20 + 0.1t^2$.
Acceleration $a$ is defined as the rate of change of velocity with respect to time,$a = \frac{dv}{dt}$.
$a = \frac{d}{dt}(20 + 0.1t^2) = 0 + 0.1 \times 2t = 0.2t$.
Since the acceleration $a = 0.2t$ depends on time $t$,it is not constant.
Therefore,the body is undergoing non-uniform acceleration.

(A) Given displacement $x = 2t^2 + t + 5$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(2t^2 + t + 5) = 4t + 1$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(4t + 1) = 4$.
Since the acceleration is constant,at $t = 2 \ s$,the acceleration remains $4 \ m/s^2$.

(C) The velocity $v$ is the first derivative of position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2$
The acceleration $a_{acc}$ is the derivative of velocity $v$ with respect to time $t$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt$
To find the time when acceleration is zero,set $a_{acc} = 0$:
$2a - 6bt = 0$
$6bt = 2a$
$t = \frac{2a}{6b} = \frac{a}{3b}$

(C) The displacement of the body is given by the equation: $x = at + bt^2 - ct^3$.
To find the velocity $(v)$,we differentiate the displacement with respect to time $(t)$:
$v = \frac{dx}{dt} = \frac{d}{dt}(at + bt^2 - ct^3) = a + 2bt - 3ct^2$.
To find the acceleration $(a_{acc})$,we differentiate the velocity with respect to time $(t)$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(a + 2bt - 3ct^2) = 0 + 2b - 6ct$.
Therefore,the acceleration of the body is $2b - 6ct$.

(C) Acceleration is defined as the rate of change of velocity with respect to time,given by $\vec{a} = \frac{d\vec{v}}{dt}$.
Since velocity $\vec{v}$ is a vector quantity,it has both magnitude (speed) and direction.
If the magnitude of velocity changes,the particle undergoes acceleration.
If the direction of velocity changes,the particle also undergoes acceleration (e.g.,in uniform circular motion).
Therefore,if either the magnitude or the direction of velocity changes,the acceleration of the particle changes.
Thus,the correct option is $(c)$.

(D) The displacement is given by $s = 3t^3 + 7t^2 + 14t + 8$.
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 8) = 9t^2 + 14t + 14$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(9t^2 + 14t + 14) = 18t + 14$.
At time $t = 1 \ s$,the acceleration is:
$a = 18(1) + 14 = 18 + 14 = 32 \ m/s^2$.

(C) The instantaneous velocity is given by $v = \frac{\Delta x}{\Delta t}$.
Using the data from the table:
For the interval $t = 0$ to $t = 1 \ s$: $v_1 = \frac{0 - (-2)}{1} = 2 \ m/s$.
For the interval $t = 1$ to $t = 2 \ s$: $v_2 = \frac{6 - 0}{1} = 6 \ m/s$.
For the interval $t = 2$ to $t = 3 \ s$: $v_3 = \frac{16 - 6}{1} = 10 \ m/s$.
Since the velocity is changing with time $(2 \ m/s, 6 \ m/s, 10 \ m/s)$,the motion is non-uniform.
Since the velocity is increasing,the motion is accelerated. Thus,the motion is non-uniform and accelerated.

(B) The displacement of the particle is given by the equation $s = 2t^2 + 2t + 4$.
To find the velocity $v$,we differentiate the displacement $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 + 2t + 4) = 4t + 2$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(4t + 2) = 4$.
Thus,the acceleration of the particle is $4 \ m/s^2$.

(B) The area under the acceleration-time graph gives the change in velocity of the particle.
Since the particle starts from rest,its initial velocity $u = 0$.
The maximum velocity is achieved at $t = 11 \, s$,where the acceleration becomes zero.
Thus,the maximum velocity $v_{\max}$ is equal to the area of the triangle $OAB$ in the given graph.
$v_{\max} = \text{Area of } \Delta OAB = \frac{1}{2} \times \text{base} \times \text{height}$
$v_{\max} = \frac{1}{2} \times 11 \, s \times 10 \, m/s^2 = 55 \, m/s$.
Therefore,the maximum speed of the particle is $55 \, m/s$.

(B) Given:
Initial velocity $\vec{v_1} = +10 \, m/s$ (taking the direction of motion as positive).
Final velocity $\vec{v_2} = -2 \, m/s$ (since it is in the opposite direction).
Time interval $t = 4 \, s$.
Acceleration $\vec{a}$ is defined as the rate of change of velocity:
$\vec{a} = \frac{\vec{v_2} - \vec{v_1}}{t}$
Substituting the values:
$\vec{a} = \frac{-2 - 10}{4} = \frac{-12}{4} = -3 \, m/s^2$.
Thus,the acceleration produced is $-3 \, m/s^2$.

(D) Average acceleration is defined as the change in velocity divided by the time interval.
$a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$
Given $v(t) = 10 + 2t^2$,$t_1 = 2 \ s$,and $t_2 = 5 \ s$.
Calculate velocity at $t_1 = 2 \ s$: $v(2) = 10 + 2(2)^2 = 10 + 8 = 18 \ m/s$.
Calculate velocity at $t_2 = 5 \ s$: $v(5) = 10 + 2(5)^2 = 10 + 50 = 60 \ m/s$.
Now,calculate average acceleration:
$a_{avg} = \frac{60 - 18}{5 - 2} = \frac{42}{3} = 14 \ m/s^2$.

(B) The relationship between acceleration $a$ and velocity $v$ is given by $a = v \frac{dv}{dx}$,which implies $\int_{u}^{v} v \ dv = \int_{0}^{x} a \ dx$.
This means $\frac{v^2 - u^2}{2} = \text{Area under the } a-x \text{ graph}$.
Given $u = 0.8 \ m/s$ at $x = 0$.
The area under the $a-x$ graph from $x = 0$ to $x = 1.4$ consists of a rectangle and a trapezoid:
Area $1$ (from $x=0$ to $x=0.4$): $0.4 \times 0.4 = 0.16 \ m^2/s^2$.
Area $2$ (from $x=0.4$ to $x=0.8$): $\frac{1}{2} \times (0.4 + 0.2) \times 0.4 = 0.12 \ m^2/s^2$.
Area $3$ (from $x=0.8$ to $x=1.4$): $0.2 \times (1.4 - 0.8) = 0.2 \times 0.6 = 0.12 \ m^2/s^2$.
Total Area $= 0.16 + 0.12 + 0.12 = 0.40 \ m^2/s^2$.
Using the formula: $\frac{v^2 - (0.8)^2}{2} = 0.40$.
$v^2 - 0.64 = 0.80$.
$v^2 = 1.44$.
$v = 1.2 \ m/s$.

(C) The position of the particle is given by the equation $x(t) = a + bt - ct^2$.
To find the velocity $v(t)$,we take the first derivative of position with respect to time:
$v(t) = \frac{dx}{dt} = \frac{d}{dt}(a + bt - ct^2) = b - 2ct$.
To find the acceleration $a_{acc}(t)$,we take the derivative of velocity with respect to time:
$a_{acc}(t) = \frac{dv}{dt} = \frac{d}{dt}(b - 2ct) = -2c$.
Since $c$ is a positive constant,the acceleration is a constant negative value. Thus,the correct expression is $a_{acc} = -2c$.

(D) Speed is defined as the magnitude of velocity,i.e.,$|\vec v|$.
$1$. If $\vec v$ and $\vec a$ have the same sign (both positive or both negative),the speed of the body increases.
$2$. If $\vec v$ and $\vec a$ have opposite signs,the speed of the body decreases.
In option $(B)$,$\vec v > 0$ and $\vec a < 0$ (opposite signs),so speed decreases. This is correct.
In option $(C)$,$\vec v < 0$ and $\vec a < 0$ (same signs),so speed increases. This is correct.
Therefore,both $(B)$ and $(C)$ are correct.

(D) The acceleration $a$ of a particle is given by the relation $a = v \frac{dv}{dx}$.
Given the velocity $v = x^2 - 5x + 4$.
First,calculate the derivative of $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx}(x^2 - 5x + 4) = 2x - 5$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (x^2 - 5x + 4)(2x - 5)$.
We need to find the acceleration when the velocity $v = 0$:
$x^2 - 5x + 4 = 0$
$(x - 1)(x - 4) = 0$
So,$x = 1$ or $x = 4$.
If $x = 1$,$a = (0)(2(1) - 5) = 0$.
If $x = 4$,$a = (0)(2(4) - 5) = 0$.
Therefore,when the velocity is zero,the acceleration is $0$.

(B) The change in velocity $\Delta v$ is equal to the area under the acceleration-time $(a-t)$ graph.
The area of the triangle in the $(a-t)$ graph is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times 2\,s \times 10\,m/s^2 = 10\,m/s$
We know that $\Delta v = v_f - v_i$,where $v_f$ is the final velocity and $v_i$ is the initial velocity.
$v_f - 5\,m/s = 10\,m/s$
$v_f = 15\,m/s$
Thus,the velocity after $2\,s$ is $15\,m/s$.

(A) The displacement is given by $s = t^3 - 3t^2 + 2$.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 6t$.
The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = 6t - 6$.
Set the acceleration to zero to find the time $t$:
$6t - 6 = 0 \implies t = 1 \, s$.
Now,substitute $t = 1 \, s$ into the displacement equation to find the displacement at that instant:
$s = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0 \, m$.

(B) Given that the distance (or displacement) $s$ is proportional to $t^3$,we can write:
$s = k t^3$ (where $k$ is a constant).
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(k t^3) = 3k t^2$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(3k t^2) = 6k t$.
Since $a = 6k t$,it implies that $a \propto t$.
This represents a linear relationship passing through the origin,which corresponds to a straight line graph starting from the origin.
Looking at the provided options,the graph that shows $a$ increasing linearly with $t$ is Graph $B$.

(D) Speed increases when the velocity and acceleration have the same sign.
Given that the object has a positive velocity at $t_1$,its speed will increase if the acceleration remains positive throughout the interval from $t_1$ to $t_2$.
In graph $I$,the acceleration $a$ is positive and increasing,so the velocity continues to increase,and the speed increases.
In graph $II$,the acceleration $a$ is constant and positive,so the velocity increases linearly,and the speed increases.
In graph $III$,the acceleration $a$ is positive throughout the interval $[t_1, t_2]$,although it is decreasing. Since $a > 0$,the velocity continues to increase,and therefore the speed increases.
Thus,in all three graphs,the acceleration is positive,which means the velocity increases,and consequently,the speed increases for the entire interval.

(D) The acceleration-time $(a-t)$ graph shows three distinct regions:
$1$. In the first region,acceleration is constant and positive,which means velocity increases linearly with time (slope is constant and positive).
$2$. In the second region,acceleration is zero,which means velocity remains constant (a horizontal line on the $v-t$ graph).
$3$. In the third region,acceleration is constant and positive,but its magnitude is greater than in the first region. This means velocity increases linearly with time again,but with a steeper slope compared to the first region.
Therefore,the velocity-time graph should show a linear increase,followed by a horizontal line,and then a steeper linear increase. This matches the graph in option $(D)$.

(B) The change in velocity $\Delta v$ is equal to the area under the acceleration-time $(a-t)$ graph.
Assuming the train starts from rest $(v_0 = 0)$,the velocity at any time $t$ is given by $v(t) = \int_{0}^{t} a(t) \, dt$.
The maximum speed occurs when the acceleration changes from positive to zero,which is at $t = 8\,s$.
Area under the $a-t$ graph from $t = 0$ to $t = 8\,s$:
Area = (Area of triangle from $0$ to $4\,s$) + (Area of rectangle from $4$ to $8\,s$)
Area = $\frac{1}{2} \times 4\,s \times 5\,m/s^2 + (8\,s - 4\,s) \times 5\,m/s^2$
Area = $10\,m/s + 20\,m/s = 30\,m/s$.
Therefore,the maximum speed of the train is $30\,m/s$.

(A) The slope of the distance-time $(s-t)$ graph represents the velocity of the particle.
Initial velocity at time $t$ is $v_i = \tan(45^{\circ}) = 1 \, m/s$.
Final velocity after one second is $v_f = \tan(60^{\circ}) = \sqrt{3} \, m/s$.
The average acceleration is defined as the change in velocity divided by the time interval:
$a_{av} = \frac{v_f - v_i}{\Delta t}$.
Substituting the values:
$a_{av} = \frac{\sqrt{3} - 1}{1} = (\sqrt{3} - 1) \, m/s^2$.

(C) Given the displacement equation: $x^2 = 1 + t^2$.
Differentiating both sides with respect to $t$:
$2x \frac{dx}{dt} = 2t$
$\frac{dx}{dt} = \frac{t}{x} = v$ (velocity).
Now,differentiate velocity $v$ with respect to $t$ to find acceleration $a$:
$a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{t}{x} \right)$
Using the quotient rule $\frac{d}{dt} (u/v) = \frac{v(du/dt) - u(dv/dt)}{v^2}$:
$a = \frac{x(1) - t(\frac{dx}{dt})}{x^2}$
Substitute $\frac{dx}{dt} = \frac{t}{x}$ into the equation:
$a = \frac{x - t(\frac{t}{x})}{x^2}$
$a = \frac{x - \frac{t^2}{x}}{x^2}$
$a = \frac{x}{x^2} - \frac{t^2}{x^3}$
$a = \frac{1}{x} - \frac{t^2}{x^3}$.

(A) The speed of a particle is the magnitude of its velocity vector $v$. The rate of change of speed is given by the tangential component of acceleration,$a_t = \frac{dv}{dt} = \frac{a \cdot v}{|v|}$.
$1$. If $v \cdot a > 0$,the angle between velocity and acceleration is acute $(0^\circ \le \theta < 90^\circ)$,so the tangential acceleration is positive,and the speed increases.
$2$. If $v \cdot a < 0$,the angle between velocity and acceleration is obtuse $(90^\circ < \theta \le 180^\circ)$,so the tangential acceleration is negative,and the speed decreases.
$3$. The cross product $v \times a$ relates to the change in direction of velocity (centripetal acceleration),not the change in speed in one-dimensional motion. Thus,option $(a)$ is correct.

(A) $\frac{dv}{dt}$ is the definition of acceleration, which is the rate of change of velocity vector. Thus, $(A \rightarrow p)$.
$(B)$ $\frac{d|v|}{dt}$ represents the rate of change of the magnitude of velocity, which is the tangential acceleration or the rate of change of speed. Thus, $(B \rightarrow q)$.
$(C)$ $\frac{dr}{dt}$ is the definition of velocity, which is the rate of change of the position vector. Thus, $(C \rightarrow r)$.
$(D)$ $\left|\frac{dr}{dt}\right|$ is the magnitude of the velocity vector, which is the speed of the particle. Thus, $(D \rightarrow s)$.

(B) Given: Initial velocity $v_0 < 0$ and acceleration $a > 0$ (constant).
$(A)$ Velocity-time graph: $v(t) = v_0 + at$. The slope is $\frac{dv}{dt} = a > 0$ (positive). As $v$ goes from negative to zero,its magnitude $|v|$ decreases. Thus,$(A) \rightarrow Q, T$.
$(B)$ Acceleration-time graph: Since acceleration is constant and positive,the slope of the $a-t$ graph is zero. Thus,$(B) \rightarrow R, U$.
$(C)$ Displacement-time graph: $s(t) = s_0 + v_0t + \frac{1}{2}at^2$. The slope is $\frac{ds}{dt} = v(t)$. Since $v$ is negative,the slope is negative. As $t$ increases,$v$ becomes less negative (approaching zero),so the magnitude of the slope $|v|$ decreases. Thus,$(C) \rightarrow P, T$.
Therefore,the correct matching is $(A) \rightarrow Q, T; (B) \rightarrow R, U; (C) \rightarrow P, T$. Note: Option $(B)$ is the closest match provided in the choices.

(A) Retardation is defined as the negative acceleration,which means the acceleration vector is directed opposite to the velocity vector. This causes a decrease in the magnitude of velocity (speed) over time.
The $Assertion$ is correct because retardation,by definition,acts in the direction opposite to the velocity to reduce it.
The $Reason$ is also correct because retardation is indeed the time rate of decrease of speed.
Since the decrease in speed is caused by the acceleration acting opposite to the velocity,the $Reason$ correctly explains why retardation is defined as it is in the $Assertion$.

(N/A) The given $a-t$ graph shows that initially,the body is moving with a constant velocity (acceleration is zero).
Then,for a short interval of time,the acceleration increases to a maximum value and then decreases back to zero.
Finally,the body continues to move with a constant velocity again.
$A$ suitable physical situation for this graph is a hammer moving with a uniform velocity striking a nail. During the impact,the hammer experiences a brief period of high acceleration (deceleration) before continuing its motion or coming to rest.

(N/A) The time rate of change of velocity is called acceleration.
Let a particle be moving in a straight line and at time $t_{1}$ and $t_{2}$ its velocities are $v_{1}$ and $v_{2}$ respectively.
Thus,the change in velocity of the particle in time interval $\Delta t = t_{2} - t_{1}$ is $\Delta v = v_{2} - v_{1}$.
According to the definition of average acceleration:
$\text{Average acceleration} = \frac{\text{change in velocity}}{\text{time interval}}$
$\langle a \rangle = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{\Delta v}{\Delta t}$
Average acceleration is a vector quantity,and its direction is the same as the direction of the change in velocity $(\Delta v)$.
The $SI$ unit of acceleration is $m/s^{2}$.
To understand how the velocity changes at a specific instant,we define instantaneous acceleration by taking the limit $\Delta t \rightarrow 0$:
$a = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$
Since velocity $v = \frac{dx}{dt}$,we can write acceleration as the second derivative of position with respect to time:
$a = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^{2}x}{dt^{2}}$
If $\frac{dv}{dt} > 0$,the acceleration is in the direction of the positive $X$-axis,and if $\frac{dv}{dt} < 0$,the acceleration is in the direction of the negative $X$-axis.

(N/A) Velocity is a vector quantity,meaning it has both magnitude (speed) and direction. Therefore,velocity can be changed in the following ways:
$(1)$ By changing only the magnitude of velocity (speed).
$(2)$ By changing only the direction of velocity.
$(3)$ By changing both the magnitude and the direction of velocity.
Since acceleration is defined as the rate of change of velocity,any of these changes will result in the production of acceleration.

(N/A) The $x-t$ (position-time) graph represents the motion of an object. The slope of the $x-t$ graph gives the velocity of the object. The curvature of the graph indicates the acceleration.
$(a)$ For positive acceleration $(a > 0)$: The graph is concave upwards (like a parabola opening upwards). The velocity increases with time.
$(b)$ For negative acceleration $(a < 0)$: The graph is concave downwards (like a parabola opening downwards). The velocity decreases with time.
$(c)$ For zero acceleration $(a = 0)$: The graph is a straight line with a constant slope,indicating constant velocity.

(N/A) Acceleration is defined as the time rate of change of velocity.
Average acceleration is the change in velocity divided by the time interval during which the change occurs. If a particle has velocities $v_{1}$ and $v_{2}$ at times $t_{1}$ and $t_{2}$ respectively,the average acceleration $\langle a \rangle$ is given by:
$\langle a \rangle = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{\Delta v}{\Delta t}$
Average acceleration is a vector quantity,and its direction is the same as the direction of the change in velocity $\Delta v$.
Instantaneous acceleration $a$ is defined as the limit of the average acceleration as the time interval $\Delta t$ approaches zero:
$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$
It represents the acceleration of an object at a specific instant of time.

(A) When the velocity (speed in a specific direction) and acceleration have the same sign,the acceleration acts in the direction of motion.
If both are positive,the object is moving in the positive direction and accelerating in the same direction,causing the speed to increase.
If both are negative,the object is moving in the negative direction and accelerating in the negative direction,which also causes the magnitude of the velocity (speed) to increase.
Therefore,in both cases,the speed of the object increases.

(B) When an object is moving,its velocity vector points in the direction of motion.
If the speed of the object is decreasing,it means the object is undergoing retardation or deceleration.
In physics,deceleration occurs when the acceleration vector acts in the direction opposite to the velocity vector.
Therefore,for a moving object with decreasing speed,the direction of acceleration is opposite to the direction of velocity.

(N/A) When the velocity of an object decreases with respect to time,it is termed as retardation or deceleration.
In reality,negative acceleration is commonly called retardation.
Its $SI$ unit is $m/s^2$.

(D) Acceleration is defined as the rate of change of velocity with respect to time,given by $a = \frac{dv}{dt}$.
An increase in velocity does not necessarily imply an increase in acceleration.
If the velocity increases at a constant rate,the acceleration remains constant (e.g.,uniform acceleration).
If the velocity increases at an increasing rate,the acceleration increases.
If the velocity increases at a decreasing rate,the acceleration decreases.
Therefore,the change in acceleration depends entirely on the nature of the motion,specifically how the velocity changes over time.

(A) The speed of a particle increases if the velocity and acceleration are in the same direction. If the velocity is positive,a positive acceleration $(a > 0)$ increases the speed. Thus,$(1)$ matches with $(b)$.
The speed of a particle decreases if the velocity and acceleration are in opposite directions. If the velocity is positive,a negative acceleration $(a < 0)$ decreases the speed. Thus,$(2)$ matches with $(a)$.
Therefore,the correct matching is $1-b$ and $2-a$.

(N/A) Acceleration is defined as the rate of change of velocity with respect to time.
Its direction is the same as the direction of the change in velocity.
The $SI$ unit of acceleration is $m/s^2$ or $m s^{-2}$.

(N/A) Retardation is defined as the time rate of decrease in velocity or speed. Its direction is always opposite to the direction of velocity or speed.

(B) particle experiences retardation or deceleration when its velocity and acceleration vectors point in opposite directions.
Mathematically,if the velocity $v$ and acceleration $a$ have opposite signs (i.e.,$v \cdot a < 0$),the speed of the particle decreases over time,which is defined as retardation.

(B) In one-dimensional motion,if the velocity $(v)$ and acceleration $(a)$ are in opposite directions,the acceleration acts as a retardation (deceleration).
This means the acceleration opposes the motion of the object.
As a result,the magnitude of the velocity decreases over time.
Therefore,the correct answer is that the magnitude of velocity decreases.

(A) Let the position vector of a particle be $\vec{r}(t)$.
$1$. The first differentiation of the position vector with respect to time $t$ gives the velocity vector:
$\vec{v} = \frac{d\vec{r}}{dt}$
$2$. The second differentiation of the position vector with respect to time $t$ (or the first differentiation of velocity) gives the acceleration vector:
$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$
Therefore,the first differentiation gives velocity and the second differentiation gives acceleration.

(N/A) For motion in a straight line,the angle between velocity and acceleration can only be $0^{\circ}$ or $180^{\circ}$.
$1$. If the angle is $0^{\circ}$,the velocity and acceleration are in the same direction,which means the speed of the object increases (accelerated motion).
Example: $A$ car starting from rest and speeding up in a straight path.
$2$. If the angle is $180^{\circ}$,the velocity and acceleration are in opposite directions,which means the speed of the object decreases (retarded motion).
Example: $A$ car applying brakes while moving in a straight line.

(A) The average acceleration over a time interval $\Delta t$ is defined as $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$.
The instantaneous acceleration is defined as $\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt}$.
If the acceleration is constant,then the rate of change of velocity is uniform at every instant.
Therefore,the average acceleration over any time interval is equal to the instantaneous acceleration at any point within that interval.

(B) For a particle moving with constant acceleration $a$:
$1$. The acceleration-time graph is a horizontal line,as $a$ is constant.
$2$. The velocity-time relation is given by $v(t) = u + at$,which represents a straight line with a non-zero slope.
$3$. The position-time relation is given by $x(t) = x_0 + ut + \frac{1}{2}at^2$,which represents a parabola.
Comparing these characteristics with the given options,the second set of graphs (represented by image $981-$b614) correctly shows a constant acceleration,a linearly increasing velocity,and a parabolic position-time graph.

(A) The velocity $v$ is given as a function of position $s$,where the rate of change of velocity with respect to position is $\frac{dv}{ds} = 5\,m^{-1}$.
We know that acceleration $a$ is defined as $a = v \frac{dv}{ds}$.
Given $v = 20\,m/s$ and $\frac{dv}{ds} = 5\,m^{-1}$.
Substituting these values,we get $a = 20 \times 5 = 100\,m/s^2$.

(A) The magnitude of velocity is defined as speed.
Mathematically,$v = |\vec{v}|$,where $v$ is speed and $\vec{v}$ is velocity.
If the speed $v = 0$ at an instant,then the magnitude of the velocity must also be zero.
Therefore,the velocity $\vec{v}$ must be zero at that instant.
Note that the acceleration does not necessarily have to be zero; for example,a ball thrown vertically upward has zero velocity and zero speed at its highest point,but it still experiences a constant acceleration due to gravity $(g = 9.8 \ m/s^2)$.

(B) If the speed of a body is increasing,the acceleration must be in the same direction as the velocity.
However,the magnitude of acceleration does not necessarily have to be constant or positive.
For example,if a particle moves in the positive direction with velocity $v > 0$,and its acceleration $a$ is positive and decreasing (e.g.,$a = 2 - t$),the speed will still increase as long as $a > 0$.
Similarly,if a particle moves in the negative direction with velocity $v < 0$,and its acceleration $a$ is negative and its magnitude is decreasing (e.g.,$a = -2 + t$),the speed $|v|$ will still increase.
Therefore,the acceleration can be positive or negative,and its magnitude can be decreasing while the speed is increasing.

(A) The speed of a particle is defined as the magnitude of its velocity,$|v|$.
For the speed to increase,the acceleration must act in the same direction as the velocity.
If the velocity $v$ is positive $(v > 0)$,the acceleration $a$ must also be positive $(a > 0)$ to increase the speed.
If the velocity $v$ is negative $(v < 0)$,the acceleration $a$ must also be negative $(a < 0)$ to increase the magnitude of the velocity.
Therefore,the speed increases when the product of velocity and acceleration is positive,i.e.,$v \cdot a > 0$.
Among the given options,$v > 0, a > 0$ (Option $A$) and $v < 0, a < 0$ (Option $B$) satisfy this condition. However,typically in such multiple-choice questions,the primary condition is that they have the same sign.

(B) The acceleration $a$ is defined as the rate of change of velocity with respect to time,$a = \frac{dv}{dt}$.
Using the chain rule,we can express this in terms of position $x$:
$a = \frac{dv}{dx} \cdot \frac{dx}{dt}$.
Since velocity $v = \frac{dx}{dt}$,we substitute this into the equation:
$a = \frac{dv}{dx} \cdot v = v \frac{dv}{dx}$.
Therefore,the correct option is $B$.