Acceleration Questions in English

(N/A) $(i)$ $A$ body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
$(ii)$ $A$ body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant,meaning its velocity changes by unequal amounts in equal intervals of time.

(B) Initial speed of the bus $(u)$ = $80 \times \frac{5}{18} = 22.22 \, m/s$.
Final speed of the bus $(v)$ = $60 \times \frac{5}{18} = 16.66 \, m/s$.
Time taken $(t)$ = $5 \, s$.
Acceleration $(a)$ is given by the formula $a = \frac{v - u}{t}$.
Substituting the values: $a = \frac{16.66 - 22.22}{5} = \frac{-5.56}{5} = -1.112 \, m/s^2$.
Thus,the acceleration of the bus is $-1.112 \, m/s^2$.

(C) Initial velocity of the train,$u = 0 \, m/s$.
Final velocity of the train,$v = 40 \, km/h = 40 \times \frac{5}{18} \, m/s = 11.11 \, m/s$.
Time taken,$t = 10 \, min = 10 \times 60 \, s = 600 \, s$.
Using the formula for acceleration,$a = \frac{v - u}{t}$.
Substituting the values,$a = \frac{11.11 - 0}{600} = 0.0185 \, m/s^{2}$.
Hence,the acceleration of the train is $0.0185 \, m/s^{2}$.

(D) Given that displacement $s$ is proportional to the square of time $t$,we can write $s = kt^2$,where $k$ is a constant.
Velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$.
Acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.
Since $2k$ is a constant,the acceleration of the object is constant or uniform.
Therefore,the object moves with uniform acceleration.

(B) The slope of a velocity-time graph is defined as the change in velocity divided by the change in time.
Mathematically, $\text{Slope} = \frac{\Delta v}{\Delta t}$.
Since acceleration is defined as the rate of change of velocity $(a = \frac{dv}{dt})$, the slope of the velocity-time graph represents the acceleration of the object.
Therefore, the correct option is $B$.

(B) Acceleration is defined as the rate of change of velocity with respect to time.
$\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}}$
The $SI$ unit of velocity is $\text{meters per second}$ ($m/s$ or $m s^{-1}$) and the $SI$ unit of time is $\text{second}$ $(s)$.
Substituting these units into the formula:
$\text{Unit of acceleration} = \frac{m s^{-1}}{s} = m s^{-2}$
Therefore,the $(second)^{2}$ appears in the denominator because velocity is divided by time,resulting in the square of the time unit in the denominator.

(B) The displacement $s$ is proportional to the square of time $t$,which can be written as $s \propto t^2$ or $s = kt^2$,where $k$ is a constant.
According to the equation of motion $s = ut + \frac{1}{2}at^2$,if the initial velocity $u = 0$,then $s = \frac{1}{2}at^2$.
Comparing $s = kt^2$ with $s = \frac{1}{2}at^2$,we get $k = \frac{1}{2}a$,which implies $a = 2k$.
Since $k$ is a constant,the acceleration $a$ is also constant.
Therefore,the body possesses uniform acceleration.

(C) The slope of a velocity$-$time graph is defined as the change in velocity divided by the change in time.
Mathematically, $\text{Slope} = \frac{\Delta v}{\Delta t}$.
Since the rate of change of velocity with respect to time is defined as acceleration, the slope of the velocity$-$time graph represents the acceleration of the object.

(N/A) $(i)$ The rate of change of velocity with respect to time is called acceleration.
$(ii)$ Acceleration is considered positive if the velocity of an object increases in the direction of motion.
$(iii)$ Acceleration is considered negative (also known as retardation or deceleration) if the velocity of an object decreases,meaning the acceleration is in the direction opposite to the direction of motion.

(N/A) $(i)$ If the velocity of a body is increasing with respect to time,the acceleration is said to be positive.
$(ii)$ If the velocity of a body is decreasing with respect to time,the acceleration is said to be negative.

(N/A) Acceleration is defined as the rate of change of velocity of an object with respect to time. Mathematically,it is expressed as $a = (v - u) / t$.
The relationship connecting initial velocity $(u)$,final velocity $(v)$,acceleration $(a)$,and time $(t)$ is given by the first equation of motion: $v = u + at$.
An example of motion with uniform acceleration is the motion of a freely falling body under the influence of gravity.

(N/A) $(i)$ Positive acceleration: This occurs when the velocity of an object increases with time. Example: $A$ car accelerating from rest to gain speed on a straight road.
$(ii)$ Negative acceleration (Retardation/Deceleration): This occurs when the velocity of an object decreases with time. Example: Applying brakes to a moving car,causing it to slow down.
$(iii)$ Zero acceleration: This occurs when an object moves with a constant velocity (uniform motion),meaning there is no change in speed or direction. Example: $A$ car moving at a constant speed of $60 \ km/h$ on a straight,level road.

(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 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.

(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.

(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.

(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.