Mix Examples - MOTION Questions in English

(N/A) Case $1$: Initial velocity $u = 72 \ km \ h^{-1} = 20 \ m \ s^{-1}$; Final velocity $v = 36 \ km \ h^{-1} = 10 \ m \ s^{-1}$; Distance $S = 25 \ m$.
Using the equation of motion $v^{2} - u^{2} = 2aS$:
$(10)^{2} - (20)^{2} = 2 \times a \times 25$
$100 - 400 = 50a$
$-300 = 50a$
$a = -6 \ m \ s^{-2}$.
Case $2$: The car starts from $u = 72 \ km \ h^{-1} = 20 \ m \ s^{-1}$ and comes to rest $(v = 0)$ with the same retardation $a = -6 \ m \ s^{-2}$.
$(i)$ To find total time $t$,use $v = u + at$:
$0 = 20 + (-6)t$
$6t = 20$
$t = 3.33 \ s$.
$(ii)$ To find total distance $S$,use $v^{2} - u^{2} = 2aS$:
$0^{2} - (20)^{2} = 2 \times (-6) \times S$
$-400 = -12S$
$S = 400 / 12 = 33.33 \ m$.

(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.

(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 $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10\, s \times 10\, m\, s^{-1} = 50\, m$.

(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,$v^2 - u^2 = 2aS$.
$(30)^2 - (5)^2 = 2 \times 5 \times S$
$900 - 25 = 10S$
$875 = 10S$
$S = 87.5\, m$.

(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.

(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 = $\frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0\, km}{3.3\, h} = 0\, km\, h^{-1}$.

(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 $1$.

(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 $False$.

(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 $False$.
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 $t$ $(v \propto t)$.

(N/A) Given initial velocity $u = 40 \, m s^{-1}$,acceleration due to gravity $g = -10 \, m s^{-2}$ (acting downwards),and final velocity $v = 0 \, m s^{-1}$ at the maximum height.
Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - (40)^2 = 2 \times (-10) \times h$
$-1600 = -20 \times h$
$h = 80 \, m$.
Therefore,the maximum height reached is $80 \, m$.
The total distance covered is the sum of the upward and downward journey: $80 \, m + 80 \, m = 160 \, m$.
The net displacement is the difference between the final and initial position. Since the stone returns to the starting point,the net displacement is $0 \, m$.

(C) When a bomb is released from an aircraft moving with a constant horizontal velocity, it possesses the same horizontal velocity as the aircraft at the moment of release.
Due to gravity, it experiences a constant downward acceleration $(g)$.
The combination of uniform horizontal motion and uniformly accelerated vertical motion results in a curved path known as a $parabola$.
Therefore, the trajectory of the bomb is a parabola.