Mix Example - MOTION Questions in English

(A) The time taken for one revolution is $t = 1 \text{ hour } 30 \text{ minutes} = 90 \text{ minutes}$.
Converting this into seconds: $t = 90 \times 60 = 5400 \text{ s}$.
The angle covered in one complete revolution is $\theta = 2\pi \text{ radians}$.
The formula for angular speed is $\omega = \frac{\theta}{t}$.
Substituting the values: $\omega = \frac{2\pi}{5400} = \frac{\pi}{2700} \text{ rad/s}$.
Calculating the numerical value: $\omega \approx \frac{3.14159}{2700} \approx 1.1635 \times 10^{-3} \text{ rad/s}$.
Thus,the angular speed is approximately $1.16 \times 10^{-3} \text{ rad/s}$.

(N/A) Average speed is given by the formula:
$V_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$
The distance travelled by the vehicle is:
$1096 \text{ km} - 1048 \text{ km} = 48 \text{ km} = 48000 \text{ m}$
The time taken is:
$40 \text{ minutes} = 40 \times 60 \text{ s} = 2400 \text{ s}$
Therefore,the average speed is:
$V_{av} = \frac{48000 \text{ m}}{2400 \text{ s}} = 20 \text{ m/s}$
The speedometer measures the instantaneous speed of the vehicle at any given moment,not the average speed over a period of time. Therefore,the speedometer will not show this calculated average velocity.

(N/A) Given: Speed of the bus $= 60 \, km/h = 60 \times (5/18) \, m/s = 16.67 \, m/s$.
$(i)$ For a normal reaction time of $1/15 \, s$:
Distance $=$ Speed $\times$ Time $= 16.67 \times (1/15) = 1.11 \, m$.
$(ii)$ For a reaction time of $1/2 \, s$ under the influence of alcohol:
Distance $=$ Speed $\times$ Time $= 16.67 \times 0.5 = 8.335 \, m$ (approximately $8.34 \, m$).

(A) Given: Acceleration $a = -6 \, m s^{-2}$ (negative because it is in the opposite direction of motion),time $t = 2 \, s$,and final velocity $v = 0 \, m s^{-1}$ (since the car stops).
$(i)$ First,find the initial velocity $u$ using the first equation of motion: $v = u + at$.
$0 = u + (-6) \times 2$
$u = 12 \, m s^{-1}$.
$(ii)$ Now,calculate the distance $S$ using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
$S = (12 \times 2) + \frac{1}{2} \times (-6) \times (2)^2$
$S = 24 - 12 = 12 \, m$.
Therefore,the car travels $12 \, m$ before stopping.

(A) Given:
Radius $r = 42,250 \, km = 42,250,000 \, m$
Time period $T = 24 \, \text{hours} = 24 \times 60 \times 60 \, s = 86,400 \, s$
The linear velocity $v$ of an object moving in a circular path is given by the formula:
$v = \frac{2 \pi r}{T}$
Substituting the values:
$v = \frac{2 \times 3.14159 \times 42,250 \, km}{86,400 \, s}$
$v \approx \frac{265,464}{86,400} \, km/s$
$v \approx 3.07 \, km/s$
Thus, the linear velocity of the satellite is approximately $3.07 \, km/s$.

(N/A) Distance travelled by the cyclist $=$ length of the path $AB$ (semicircle) $=$ half the circumference of the circle.
$S = \frac{314}{2} = 157 \ m$.
$(b)$ The displacement of the cyclist is equal to the diameter $AB$ of the circle.
Since circumference $= 2 \pi r = 314 \ m$,therefore,
$r = \frac{\text{circumference}}{2 \pi} = \frac{314}{2 \times 3.14} = 50 \ m$.
Therefore,displacement of the cyclist $= 2 \times r = 2 \times 50 = 100 \ m$ towards south.
$(c)$ Average velocity is defined as the total displacement divided by the total time taken.
First,find the time taken: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{157 \ m}{15.7 \ m s^{-1}} = 10 \ s$.
Average velocity $= \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{100 \ m}{10 \ s} = 10 \ m s^{-1}$ towards south.

(B) Initial velocity $u = 80 \, km \, h^{-1} = 80 \times \frac{5}{18} = \frac{200}{9} \, m \, s^{-1} \approx 22.22 \, m \, s^{-1}$.
Final velocity $v = 50 \, km \, h^{-1} = 50 \times \frac{5}{18} = \frac{125}{9} \, m \, s^{-1} \approx 13.89 \, m \, s^{-1}$.
Time $t = 4 \, s$.
Using the formula for acceleration $a = \frac{v - u}{t}$:
$a = \frac{\frac{125}{9} - \frac{200}{9}}{4} = \frac{-75}{9 \times 4} = \frac{-75}{36} = -2.083 \, m \, s^{-2}$.
The negative sign indicates retardation or deceleration.

(A) Given velocity $v = 120 \, km \, h^{-1}$.
Time $t = 30 \, s$.
First,convert time into hours: $t = 30 \, s = 30 / 3600 \, h = 1 / 120 \, h$.
Using the formula for distance: $S = v \times t$.
$S = 120 \, km \, h^{-1} \times (1 / 120) \, h = 1 \, km$.
Therefore,the train will cover a distance of $1 \, km$ in $30 \, s$.

(N/A) For train $A$:
Initial velocity $u_A = 54\, km\, h^{-1} = 54 \times \frac{5}{18} = 15\, m\, s^{-1}$.
Final velocity $v_A = 0\, m\, s^{-1}$.
Time taken $t_A = 5\, s$.
Distance travelled by train $A$ is the area under the velocity-time graph $AB$:
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5\, s \times 15\, m\, s^{-1} = 37.5\, m$.
For train $B$:
Initial velocity $u_B = 36\, km\, h^{-1} = 36 \times \frac{5}{18} = 10\, m\, s^{-1}$.
Final velocity $v_B = 0\, m\, s^{-1}$.
Time taken $t_B = 10\, s$.
Distance travelled by train $B$ is the area under the velocity-time graph $CD$:
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10\, s \times 10\, m\, s^{-1} = 50\, m$.

(66 S) Length of the train $= 100 \, m = 0.1 \, km$.
Length of the bridge $= 1 \, km$.
Total distance to be covered to cross the bridge $= \text{Length of train} + \text{Length of bridge} = 0.1 \, km + 1 \, km = 1.1 \, km$.
Velocity of the train $= 60 \, km/h$.
Time taken $(t) = \frac{\text{Total distance}}{\text{Velocity}} = \frac{1.1 \, km}{60 \, km/h} = \frac{1.1}{60} \, h$.
To convert time into seconds, multiply by $3600 \, s/h$:
$t = \frac{1.1}{60} \times 3600 \, s = 1.1 \times 60 \, s = 66 \, s$.

(DISTANCE = 231 M, DISPLACEMENT = 49 M) Given: Diameter $d = 49 \,m$,Time for one round $T = 20 \,s$,Total time $t = 30 \,s$.
First,calculate the number of rounds completed in $30 \,s$:
$n = \frac{t}{T} = \frac{30}{20} = 1.5 \, \text{rounds}$.
Distance covered is the total path length:
$S = n \times (2 \pi r) = n \times (\pi d) = 1.5 \times \frac{22}{7} \times 49 = 1.5 \times 22 \times 7 = 231 \,m$.
Displacement is the shortest distance between the initial and final positions:
After $1.5$ rounds,the athlete is at the point diametrically opposite to the starting point.
Therefore,the displacement is equal to the diameter of the circular track,which is $49 \,m$.

(A) Let the distance between Ali's home and school be $x$.
During the trip to school:
Distance $= x$,Speed $= 20 \, km \, h^{-1}$.
Time taken $t_1 = \frac{x}{20} \, h$.
During the return trip:
Distance $= x$,Speed $= 30 \, km \, h^{-1}$.
Time taken $t_2 = \frac{x}{30} \, h$.
Average speed is defined as the total distance divided by the total time taken:
Average Speed $= \frac{\text{Total Distance}}{\text{Total Time}} = \frac{x + x}{t_1 + t_2} = \frac{2x}{\frac{x}{20} + \frac{x}{30}}$.
Calculating the denominator:
$\frac{x}{20} + \frac{x}{30} = \frac{3x + 2x}{60} = \frac{5x}{60} = \frac{x}{12}$.
Therefore,Average Speed $= \frac{2x}{x/12} = 2x \times \frac{12}{x} = 24 \, km \, h^{-1}$.

(N/A) Given: Initial velocity $u = 5\, m s^{-1}$,Final velocity $v = 30\, m s^{-1}$,Time $t = 5\, s$.
$(i)$ To calculate acceleration $(a)$: Using the first equation of motion,$v = u + at$.
$30 = 5 + a \times 5$
$30 - 5 = 5a$
$25 = 5a$
$a = 5\, m s^{-2}$.
$(ii)$ To calculate distance covered $(S)$: Using the third equation of motion,$v^2 - u^2 = 2aS$.
$(30)^2 - (5)^2 = 2 \times 5 \times S$
$900 - 25 = 10S$
$875 = 10S$
$S = 87.5\, m$.

(A) Given: Initial velocity $u = 36 \, km \, h^{-1} = 36 \times \frac{5}{18} = 10 \, m \, s^{-1}$.
Final velocity $v = 0 \, m \, s^{-1}$ (since the cycle stops).
Time $t = 2 \, s$.
$(i)$ To calculate retardation:
Using the first equation of motion,$v = u + at$:
$0 = 10 + a \times 2$
$2a = -10$
$a = -5 \, m \, s^{-2}$.
Retardation is the negative of acceleration,so retardation $= -(-5 \, m \, s^{-2}) = 5 \, m \, s^{-2}$.
$(ii)$ To calculate distance covered $(S)$:
Using the third equation of motion,$v^2 - u^2 = 2aS$:
$0^2 - (10)^2 = 2 \times (-5) \times S$
$-100 = -10 \times S$
$S = \frac{-100}{-10} = 10 \, m$.

(N/A) The speed-time graph is a straight line starting from $(0, 5)$ to $(50, 30)$.
$(i)$ Acceleration is the slope of the speed-time graph.
Slope $= \frac{\text{Change in speed}}{\text{Change in time}} = \frac{30 - 5}{50 - 0} = \frac{25}{50} = 0.5 \ m s^{-2}$.
$(ii)$ The distance travelled is the area under the speed-time graph.
This area forms a trapezium with parallel sides of length $5 \ m s^{-1}$ and $30 \ m s^{-1}$, and height (time) of $50 \ s$.
Area $= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Area $= \frac{1}{2} \times (5 + 30) \times 50 = \frac{1}{2} \times 35 \times 50 = 35 \times 25 = 875 \ m$.

(N/A) $(i)$ Given: Initial velocity $u = 0 \, m/s$,Distance $S = 64 \, m$,Time $t = 4 \, s$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$64 = 0(4) + \frac{1}{2} \times a \times (4)^2$
$64 = 8a$
$a = 8 \, m/s^2$.
Using the equation $v = u + at$:
$v = 0 + 8 \times 4 = 32 \, m/s$.
$(ii)$ Half the total distance is $S' = \frac{64}{2} = 32 \, m$.
Using $S' = ut + \frac{1}{2}at^2$:
$32 = 0(t) + \frac{1}{2} \times 8 \times t^2$
$32 = 4t^2$
$t^2 = 8$
$t = \sqrt{8} \approx 2.83 \, s$.

(A) Dynamics is the branch of mechanics that deals with the motion of objects and the forces that cause this motion.
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion.
Statics is the branch of mechanics that deals with objects at rest or in equilibrium under the action of forces.
Therefore,the correct branch that considers the cause of motion is dynamics.

(C) Uniform acceleration is defined as the constant rate of change of velocity with respect to time $(a = \frac{dv}{dt} = \text{constant})$.
$Mr. Y$'s statement,"the longer you go, the faster you go," refers to the time duration $(t)$. Since $v = u + at$, as time $(t)$ increases, velocity $(v)$ increases linearly, which correctly describes uniform acceleration.
$Mr. X$'s statement,"the farther you go, the faster you go," refers to the displacement $(s)$. Using the equation $v^2 = u^2 + 2as$, we can see that as displacement $(s)$ increases, velocity $(v)$ also increases. Therefore, both statements are physically correct descriptions of uniform acceleration.

(C) For a body starting from rest $(u = 0)$ and moving with constant acceleration $(a)$ along an inclined plane, the distance $(s)$ covered in time $(t)$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.
Since $u = 0$, the equation simplifies to $s = \frac{1}{2}at^2$, which implies $s \propto t^2$.
Let the total distance be $S$ and the total time taken be $T = 4$ seconds.
We are asked to find the time $(t')$ taken to cover one-fourth of the distance, i.e., $s' = \frac{S}{4}$.
Using the proportionality $s \propto t^2$, we have the ratio: $\frac{s'}{S} = \frac{(t')^2}{T^2}$.
Substituting the values: $\frac{S/4}{S} = \frac{(t')^2}{4^2}$.
$\frac{1}{4} = \frac{(t')^2}{16}$.
$(t')^2 = \frac{16}{4} = 4$.
$t' = \sqrt{4} = 2$ seconds.
Therefore, the body covers one-fourth of the distance in $2$ seconds.

(D) Average velocity is defined as the total displacement divided by the total time taken.
Displacement is a vector quantity representing the shortest distance between the initial and final positions.
In this problem,the car starts at a point,travels $100\, km$ east,$100\, km$ south,and then returns to the exact same starting point.
Since the final position is the same as the initial position,the total displacement is $0\, km$.
Therefore,the average velocity = $\frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0\, km}{3.3\, h} = 0\, km\, h^{-1}$.

(A) Given that the displacement $s$ is proportional to the cube of time $t$,we can write $s = kt^3$,where $k$ is a constant.
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(kt^3) = 3kt^2$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(3kt^2) = 6kt$.
Since $a = 6kt$,the acceleration $a$ is directly proportional to time $t$. Therefore,the magnitude of the acceleration increases linearly with time.

(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$.
Given,total time $T = t_1 + t_2 = 4\, s$.
The velocity-time graph for this motion is a triangle with base $T = 4\, s$ and height $v_{max} = 8\, m s^{-1}$.
The distance travelled is equal to the area under the velocity-time graph.
Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 4\, s \times 8\, m s^{-1} = 16\, m$.
Thus,the total distance travelled is $16\, m$.

(C) $1$. Distance is the total path length covered by the body. For three-quarters of a circle, the distance is $\frac{3}{4}$ of the circumference $(2\pi r)$.
Distance $= \frac{3}{4} \times 2\pi r = \frac{3\pi r}{2}$.
$2$. Displacement is the shortest straight-line distance between the initial and final positions. If the body starts at point $A$ and moves three-quarters of the circle to point $B$, the angle subtended at the center is $270^{\circ}$ (or $-90^{\circ}$).
$3$. The initial position vector is $(r, 0)$ and the final position vector is $(0, -r)$.
The displacement magnitude is the hypotenuse of a right-angled triangle with sides $r$ and $r$.
Displacement $= \sqrt{r^2 + r^2} = \sqrt{2r^2} = \sqrt{2}r$.

(B) For any motion,the distance covered is always greater than or equal to the magnitude of the displacement.
Mathematically,$\text{Distance} \geq |\text{Displacement}|$.
Therefore,the ratio of the magnitude of the displacement to the distance covered is $\frac{|\text{Displacement}|}{\text{Distance}} \leq 1$.
In a straight-line path,if the motion is unidirectional (no turning back),the distance is equal to the magnitude of the displacement,so the ratio is $1$.
If the motion involves a change in direction,the distance will be greater than the magnitude of the displacement,making the ratio less than $1$.
Thus,the ratio is always less than or equal to $1$.

(A) Let the velocity at point $A$ be $v_A = 20 \, m/s$ and at point $B$ be $v_B = 30 \, m/s$.
Let the distance between $A$ and $B$ be $s$. The velocity $v$ at the midpoint $M$ can be found using the third equation of motion: $v^2 = u^2 + 2as$.
For the segment $AM$,$v_M^2 = v_A^2 + 2a(s/2) = v_A^2 + as$.
For the segment $MB$,$v_B^2 = v_M^2 + 2a(s/2) = v_M^2 + as$.
From the first equation,$as = v_M^2 - v_A^2$. Substituting this into the second equation:
$v_B^2 = v_M^2 + (v_M^2 - v_A^2) = 2v_M^2 - v_A^2$.
Rearranging for $v_M$:
$2v_M^2 = v_A^2 + v_B^2 \implies v_M = \sqrt{\frac{v_A^2 + v_B^2}{2}}$.
Substituting the values: $v_M = \sqrt{\frac{20^2 + 30^2}{2}} = \sqrt{\frac{400 + 900}{2}} = \sqrt{\frac{1300}{2}} = \sqrt{650} = \sqrt{25 \times 26} = 5 \sqrt{26} \approx 25.49 \, m/s$.
Wait,calculating $\sqrt{650} \approx 25.495$. Let's re-check the options. $10\sqrt{6} = \sqrt{600} \approx 24.49$. Actually,the formula for velocity at the midpoint is $v_M = \sqrt{\frac{v_A^2 + v_B^2}{2}}$. Calculation: $\sqrt{650} \approx 25.5 \, m/s$.

(D) Given that the distance $s$ is directly proportional to the square of time $t$,we can write: $s \propto t^2$ or $s = kt^2$,where $k$ is a constant.
To find the velocity $v$,we differentiate the distance with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$.
To find the acceleration $a$,we differentiate the velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.
Since $k$ is a constant,$2k$ is also a constant.
Therefore,the acceleration of the body is constant.

(C) The area under a velocity$-$time graph represents the displacement or the total distance travelled by an object.
Mathematically,for a velocity $v$ and time $t$,the area is calculated as $\int v \, dt$,which corresponds to the displacement of the object over the given time interval.

(D) vector quantity is a physical quantity that has both magnitude and direction.
Displacement,velocity,and acceleration are all vector quantities because they require both magnitude and direction to be fully described.
Speed is a scalar quantity because it only represents the magnitude of the rate of change of position (distance per unit time) and does not have a specific direction associated with it.
Therefore,speed is not a vector.

(A) The average velocity $(v_{avg})$ of a body is defined as the arithmetic mean of its initial velocity $(u)$ and final velocity $(v)$ only when the acceleration $(a)$ is constant or uniform.
Mathematically,$v_{avg} = \frac{u + v}{2}$.
This formula is derived from the equations of motion under constant acceleration,specifically $v = u + at$.
Substituting $v$ in the average velocity formula: $v_{avg} = \frac{u + (u + at)}{2} = \frac{2u + at}{2} = u + \frac{1}{2}at$.
Since this relationship holds true for motion with constant acceleration,the acceleration of the body must be uniform.

(B) In a speed-time graph,the $y$-axis represents speed and the $x$-axis represents time.
If the graph is a straight line parallel to the time axis,it means that the speed of the body remains constant at all points in time.
$A$ constant speed indicates that the body is moving with uniform speed (or constant velocity if the direction is fixed).
Since the speed does not change over time,the acceleration of the body is zero.

(C) In a velocity$-$time graph,the slope represents the acceleration of the body.
If the slope is negative,it indicates that the velocity of the body is decreasing over time.
$A$ constant negative slope implies that the rate of decrease in velocity is uniform.
This phenomenon is known as uniform retardation or uniform deceleration.
Therefore,the correct option is $C$.

(D) In a distance$-$time graph,the $y$-axis represents the distance covered and the $x$-axis represents time.
If the graph is parallel to the time axis ($x$-axis),it means that as time increases,the distance covered by the body remains constant.
$A$ constant distance indicates that the body is not changing its position with respect to time.
Therefore,the body is at rest.

(A) In a distance-time graph,the slope of the line represents the speed of the object.
If the graph is a straight line inclined to the time axis,it indicates that the object covers equal distances in equal intervals of time.
This constant slope implies a constant speed.
Therefore,the body is in uniform motion.

(B) The area under a velocity$-$time graph represents the displacement of the object.
Velocity is defined as the rate of change of displacement $(v = \frac{ds}{dt})$.
Therefore,the displacement $(s)$ can be calculated by integrating the velocity with respect to time: $s = \int v \, dt$.
Geometrically,this integral corresponds to the area enclosed between the velocity$-$time curve and the time axis.
If the motion is in a straight line without changing direction,the magnitude of displacement is equal to the distance travelled. However,in general,the area under the velocity$-$time graph represents the magnitude of the displacement.

(C) The area under a speed-time graph represents the total distance covered by an object.
Since speed is a scalar quantity (distance/time),the area calculated by multiplying speed and time gives the total distance travelled.
In contrast,the area under a velocity-time graph represents the displacement of the body.
Therefore,for a speed-time graph,the correct answer is the distance travelled by the body.

(D) When an object moves in a circular path at a constant speed,its velocity changes continuously due to the change in direction at every point.
This change in velocity gives rise to an acceleration known as centripetal acceleration.
By definition,centripetal acceleration is always directed towards the centre of the circular path.
Therefore,the correct option is $D$.

(B) The first statement is $False$. Locomotion is specifically defined as the ability of an organism to move from one place to another,whereas motion is a general term for any change in position. Not all motion in animals is locomotion (e.g.,movement of internal organs).
The second statement is $False$. Mechanics is a branch of physics that deals with the motion of both living and non-living objects under the action of forces. Therefore,the overall answer is $False$.

(TRUE) The statement is True.
Kinematics is a branch of mechanics that describes the motion of objects (points,bodies,and systems of bodies) without considering the forces that cause them to move. It focuses on parameters such as position,velocity,acceleration,and time.

(FALSE) The statement is False.
Rectilinear motion is defined as the motion of an object along a straight line.
Motion along a curved path is known as curvilinear motion.
Translatory motion refers to motion where all parts of a body move in the same direction at the same time,which can occur along a straight or curved path,but it is not synonymous with rectilinear motion.

(TRUE) The statement is True.
Definition: An object is said to be at rest if its position does not change with respect to a fixed reference point (or frame of reference) as time passes.

(B) The statement is False.
Explanation:
$1$. $A$ quantity that is described completely by its magnitude alone is known as a scalar quantity (e.g.,distance,speed,mass).
$2$. $A$ vector quantity requires both magnitude and direction to be completely specified (e.g.,displacement,velocity,force).
$3$. Since the statement defines a scalar quantity as a vector quantity,it is incorrect.

(B) The statement is False.
$A$ quantity that is specified by both magnitude and direction is known as a vector quantity.
$A$ scalar quantity is a physical quantity that is specified only by its magnitude,such as distance,speed,or mass.

(B) The statement is False.
Speed is defined as the distance traveled per unit time,while velocity is defined as the displacement per unit time.
Both speed and velocity are measured in the same $SI$ unit,which is meters per second $(m/s)$.

(FALSE) The statement is False.
In one-dimensional motion with uniform velocity,the average velocity is equal to the instantaneous velocity at any point in time.
Even in non-uniform motion,the average velocity over an interval can be equal to the instantaneous velocity at a specific moment within that interval,so the statement that they are always unequal is incorrect.

(A) The statement is $True$.
Uniform motion is defined as the motion of an object in which the object travels equal distances in equal intervals of time along a straight line. Since displacement is the shortest distance between two points in a specific direction,undergoing equal displacements in equal intervals of time is the fundamental characteristic of uniform motion.

(B) The statement is $False$.
In uniform motion,an object covers equal distances in equal intervals of time,which implies a constant velocity.
For uniform motion,the position $x$ is related to time $t$ by the linear equation $x = vt + x_0$,meaning $x \propto t$.
If $x \propto t^{2}$,the motion represents constant acceleration (non-uniform motion),as the velocity $v = dx/dt$ would be proportional to $t$ $(v \propto t)$.

(TRUE) The statement is True.
Acceleration is defined as the rate of change of velocity with respect to time. Mathematically,it is expressed as $a = \frac{v - u}{t}$,where $v$ is the final velocity,$u$ is the initial velocity,and $t$ is the time taken.