Mix Example - MOTION Questions in English

(A) Displacement is defined as the shortest distance between the initial and final positions of an object.
In a circular path of radius $r$,after completing half a circle,the particle moves from one end of the diameter to the other end.
The initial position and final position are diametrically opposite.
The shortest distance between these two points is the diameter of the circle.
Therefore,displacement = $2r$.

(B) When a body is thrown vertically upward with an initial velocity $u$,it reaches its maximum height $h$ where its final velocity $v$ becomes $0$.
According to the third equation of motion: $v^2 = u^2 + 2as$.
Here,$v = 0$,$a = -g$ (acceleration due to gravity acting downwards),and $s = h$.
Substituting these values: $0^2 = u^2 + 2(-g)h$.
$0 = u^2 - 2gh$.
$2gh = u^2$.
Therefore,the maximum height reached is $h = u^2 / (2g)$.

(C) Displacement is the shortest path between the initial and final positions,while distance is the total path length covered by an object.
Since the shortest path is always less than or equal to the total path length,the magnitude of displacement is always less than or equal to the distance.
Therefore,the ratio of displacement to distance is always $\leq 1$.

(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.

(A) In a $v-t$ (velocity-time) graph,the $y$-axis represents velocity $(v)$ and the $x$-axis represents time $(t)$.
Since the graph is a straight horizontal line parallel to the time axis and above the zero mark on the $v$-axis,it indicates that the velocity of the object remains constant over time.
$A$ constant velocity implies that the object is in uniform motion.

(B) boy on a merry-go-round is performing circular motion.
In circular motion,even if the speed remains constant,the direction of motion changes at every point.
Since velocity is a vector quantity (having both magnitude and direction),a change in direction results in a change in velocity.
Therefore,the boy is in accelerated motion.

(C) The area under a velocity-time $(v-t)$ graph is calculated by multiplying velocity $(v)$ and time $(t)$.
Since $\text{velocity} = \text{displacement} / \text{time}$,the product of velocity and time gives displacement.
$\text{Area} = \text{velocity} \times \text{time} = (\text{m/s}) \times \text{s} = \text{m}$.
Therefore,the physical quantity represented by the area under a $v-t$ graph is displacement,and its $SI$ unit is meter $(m)$.

(D) In a distance-time graph,the slope of the line represents the speed of the object.
Greater the slope,higher is the speed.
Looking at the graph,the line for car $C$ has the maximum slope,followed by $D$,$A$,and $B$.
Therefore,car $C$ is the fastest and car $B$ has the minimum slope,making it the slowest car.

(A) Uniform motion is defined as the motion of an object in which it travels equal distances in equal intervals of time.
In a distance-time graph,this is represented by a straight line with a constant slope passing through the origin.
Figure $(A)$ shows a straight line starting from the origin,which indicates that the distance increases at a constant rate with respect to time,thus representing uniform motion.

(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 $B$.

(C) Distance is the total path length covered by an object,while displacement is the shortest distance between the initial and final positions.
When an object moves along a straight-line path without changing its direction,the total distance covered is equal to the magnitude of the displacement.
In cases of circular motion,to-and-fro motion,or orbital motion,the object changes its direction,causing the displacement to be less than the total distance covered.
Therefore,the correct option is $C$.

(B) No,the distance travelled by the object would not necessarily be zero.
Displacement is a vector quantity defined as the shortest distance between the initial and final positions,whereas distance is a scalar quantity representing the total path length covered by the object.
Displacement becomes zero when an object returns to its starting point (initial position = final position) after completing a journey.
However,the object has still covered a path to return to the start,meaning the total distance travelled is greater than zero.
For example,if an athlete runs one complete lap around a circular track,their displacement is $0$,but the distance travelled is equal to the circumference of the track $(2 \pi r)$.

(N/A) When an object moves with a uniform velocity,the acceleration $(a)$ is $0$ and the final velocity $(v)$ is equal to the initial velocity $(u)$,i.e.,$v = u$.
$1$. The first equation of motion,$v = u + at$,becomes $v = u + (0)t$,which simplifies to $v = u$.
$2$. The second equation of motion,$s = ut + (1/2)at^2$,becomes $s = ut + (1/2)(0)t^2$,which simplifies to $s = ut$.
$3$. The third equation of motion,$v^2 - u^2 = 2as$,becomes $v^2 - u^2 = 2(0)s$,which simplifies to $v^2 - u^2 = 0$ or $v^2 = u^2$.

(N/A) $1$. From the displacement-time graph,the velocity is the slope of the line.
$2$. For the interval $0-50 \, s$,the displacement changes from $0 \, m$ to $100 \, m$. Thus,velocity $v_1 = \frac{100 \, m - 0 \, m}{50 \, s - 0 \, s} = 2 \, m/s$.
$3$. For the interval $50-100 \, s$,the displacement changes from $100 \, m$ to $0 \, m$. Thus,velocity $v_2 = \frac{0 \, m - 100 \, m}{100 \, s - 50 \, s} = \frac{-100 \, m}{50 \, s} = -2 \, m/s$.
$4$. The velocity-time graph will show a constant velocity of $2 \, m/s$ from $0$ to $50 \, s$ and a constant velocity of $-2 \, m/s$ from $50$ to $100 \, s$.

(A) The distance travelled in the first $8 \, s$ is calculated using the equation $s = ut + \frac{1}{2}at^2$. Since the car starts from rest,$u = 0$.
$x_1 = 0 + \frac{1}{2} \times 5 \times (8)^2 = 160 \, m$.
At $t = 8 \, s$,the velocity $v$ is calculated as $v = u + at = 0 + (5 \times 8) = 40 \, m/s$.
For the remaining time,which is $12 \, s - 8 \, s = 4 \, s$,the car moves with a constant velocity of $40 \, m/s$.
The distance covered in the last $4 \, s$ is $x_2 = v \times t = 40 \times 4 = 160 \, m$.
Therefore,the total distance covered in $12 \, s$ is $x = x_1 + x_2 = 160 \, m + 160 \, m = 320 \, m$.

(D) Let the distance between $A$ and $B$ be $x\, km$.
Time taken to travel from $A$ to $B$ is $t_{1} = \frac{x}{30}\, h$.
Time taken to travel from $B$ to $A$ is $t_{2} = \frac{x}{20}\, h$.
Total distance covered $= x + x = 2x\, km$.
Total time taken $= t_{1} + t_{2} = \frac{x}{30} + \frac{x}{20} = \frac{2x + 3x}{60} = \frac{5x}{60} = \frac{x}{12}\, h$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x/12} = 2 \times 12 = 24\, km\, h^{-1}$.

(A) $(i)$ Since the velocity-time graph is a horizontal line parallel to the time axis,the velocity is constant. Therefore,the acceleration is $0\,m/s^2$.
$(ii)$ By observing the graph,the constant velocity of the cyclist is $20\,m/s$.
$(iii)$ The distance covered by the cyclist in $15\,s$ is given by the area under the velocity-time graph,which is a rectangle with height $v = 20\,m/s$ and width $t = 15\,s$.
Distance $= v \times t = 20\,m/s \times 15\,s = 300\,m$.

(N/A) When a stone is thrown vertically upwards,its initial velocity is at its maximum. As the stone rises,its velocity decreases due to the constant downward acceleration due to gravity $(g)$.
At the maximum height,the velocity of the stone becomes zero.
After reaching the maximum height,the stone begins to fall downwards. During this phase,its velocity increases in the downward direction until it hits the ground.
The velocity-time graph for this motion is a straight line with a negative slope,representing constant deceleration during the upward journey and constant acceleration during the downward journey.

(50 M) Let the initial heights be $H_1 = 150 \, m$ and $H_2 = 100 \, m$.
Both objects are dropped from rest,so their initial velocity $u = 0 \, m/s$.
Both objects fall with the same acceleration due to gravity,$g$.
Distance covered by an object in time $t$ is given by $s = ut + \frac{1}{2}gt^2$.
Since $u = 0$,the distance covered by both objects in $t = 2 \, s$ is $s = \frac{1}{2}g(2)^2 = 2g$.
After $2 \, s$,the height of the first object from the ground is $h_1 = H_1 - s = 150 - 2g$.
The height of the second object from the ground is $h_2 = H_2 - s = 100 - 2g$.
The difference in their heights after $2 \, s$ is $h_1 - h_2 = (150 - 2g) - (100 - 2g) = 150 - 100 = 50 \, m$.
Since the acceleration $g$ is the same for both,the distance covered by both objects at any time $t$ is the same $(s = \frac{1}{2}gt^2)$.
Therefore,the difference in heights at any time $t$ is $(H_1 - s) - (H_2 - s) = H_1 - H_2 = 150 - 100 = 50 \, m$.
Thus,the difference in heights remains constant and does not vary with time.

(D) Given that the object starts from rest,initial velocity $u = 0 \, m/s$.
For the first $2 \, s$,distance $s_1 = 20 \, m$ and time $t_1 = 2 \, s$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$20 = 0(2) + \frac{1}{2} a(2)^2 \Rightarrow 20 = 2a \Rightarrow a = 10 \, m/s^2$.
Since the acceleration is constant,we can calculate the velocity at any time $t$ using $v = u + at$.
For $t = 7 \, s$:
$v = 0 + (10 \times 7) = 70 \, m/s$.

(N/A) Average velocity is defined as the total change in displacement divided by the total time taken.
$1$. For the first $4 \, s$ ($t = 0$ to $4 \, s$):
Average velocity $v = \frac{\Delta x}{\Delta t} = \frac{4 - 0}{4 - 0} = \frac{4}{4} = 1 \, m/s$.
$2$. For the next $4 \, s$ ($t = 4$ to $8 \, s$):
Average velocity $v = \frac{\Delta x}{\Delta t} = \frac{4 - 4}{8 - 4} = \frac{0}{4} = 0 \, m/s$.
$3$. For the last $6 \, s$ ($t = 10$ to $16 \, s$):
Average velocity $v = \frac{\Delta x}{\Delta t} = \frac{0 - 6}{16 - 10} = \frac{-6}{6} = -1 \, m/s$.

(B) Given initial velocity,$u = 5 \times 10^{4} \, ms^{-1}$.
Acceleration,$a = 10^{4} \, ms^{-2}$.
$(i)$ Final velocity $v = 2u = 2 \times 5 \times 10^{4} = 10 \times 10^{4} \, ms^{-1}$.
Using the first equation of motion,$v = u + at$,we get $t = \frac{v - u}{a}$.
$t = \frac{10 \times 10^{4} - 5 \times 10^{4}}{10^{4}} = \frac{5 \times 10^{4}}{10^{4}} = 5 \, s$.
$(ii)$ Using the second equation of motion,$s = ut + \frac{1}{2}at^{2}$.
$s = (5 \times 10^{4}) \times 5 + \frac{1}{2} \times (10^{4}) \times (5)^{2}$.
$s = 25 \times 10^{4} + 12.5 \times 10^{4} = 37.5 \times 10^{4} \, m$.

(C) The distance travelled by an object in time $t$ with uniform acceleration $a$ and initial velocity $u$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.
First,calculate the total distance travelled in $5 \, s$:
$s_5 = u(5) + \frac{1}{2}a(5)^2 = 5u + \frac{25}{2}a$ ........... $(i)$
Next,calculate the total distance travelled in $4 \, s$:
$s_4 = u(4) + \frac{1}{2}a(4)^2 = 4u + \frac{16}{2}a = 4u + 8a$ ........... $(ii)$
The distance travelled in the interval between the $4^{th}$ and $5^{th}$ second is the difference between the distance at $5 \, s$ and $4 \, s$:
Distance $= s_5 - s_4 = (5u + \frac{25}{2}a) - (4u + 8a)$
Distance $= (5u - 4u) + (12.5a - 8a) = u + 4.5a$
Distance $= (u + \frac{9}{2}a) \, m$.

For an object moving vertically upwards,the third equation of motion is given by $v^{2} = u^{2} + 2as$.
At the maximum height,the final velocity $v$ becomes $0$.
Substituting $v = 0$,$a = -g$,and $s = h$ into the equation: $0 = u^{2} - 2gh$.
Rearranging for height,we get $h = \frac{u^{2}}{2g}$.
For the first stone with initial velocity $u_{1}$,the maximum height reached is $h_{1} = \frac{u_{1}^{2}}{2g}$.
For the second stone with initial velocity $u_{2}$,the maximum height reached is $h_{2} = \frac{u_{2}^{2}}{2g}$.
Taking the ratio of the two heights: $\frac{h_{1}}{h_{2}} = \frac{u_{1}^{2} / 2g}{u_{2}^{2} / 2g} = \frac{u_{1}^{2}}{u_{2}^{2}}$.
Therefore,the ratio of the heights reached is $h_{1}: h_{2} = u_{1}^{2}: u_{2}^{2}$.

(A) According to Newton's $Second$ $Law$ of $Motion$, the rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force.
Mathematically, $F = \frac{dp}{dt}$, where $F$ is the force and $p$ is the momentum.
Therefore, the physical quantity that corresponds to the rate of change of momentum is $Force$.

(C) If the displacement$-$time graph is a straight line parallel to the time axis,it indicates that the displacement of the body does not change with time.
Since the rate of change of displacement with respect to time is zero,the velocity of the body is zero.
Therefore,the body is at rest.

(B) In a displacement-time graph,the slope of the line represents the velocity of the body.
If the graph is parallel to the time axis,the displacement of the body does not change with time.
This means the body is at rest or stationary.
Since the body is not changing its position,its velocity is $0 \ m/s$.

(B) Displacement is defined as the shortest distance between the initial and final positions of an object.
When a body completes half a revolution on a circular path of radius $R$,it moves from one end of the diameter to the other end.
The initial position is at one end of the diameter,and the final position is at the diametrically opposite end.
The shortest distance between these two points is the diameter of the circle.
Therefore,the displacement is equal to the diameter,which is $2 \times R = 2\,R$.

(B) When a body is in motion,its position changes with respect to time.
Consequently,the distance covered by the body and its displacement from the initial point change continuously.
Therefore,distance or displacement are the physical quantities that essentially change as a body moves.

(B) The magnitude of displacement is the shortest distance between the initial and final positions of an object.
Distance is the total path length covered by the object.
When an object moves along a straight line without changing its direction,the total path length covered is exactly equal to the straight-line distance between the starting and ending points.
Therefore,the magnitude of displacement is equal to the distance travelled only when the object moves along a straight line in one direction.

(A) When an object moves along a straight path in a single direction without changing its direction, the magnitude of its displacement is equal to the total distance traveled.
Since Average Velocity = $\frac{\text{Total Displacement}}{\text{Total Time}}$ and Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$, and given that Displacement = Distance in this specific case, the ratio of average velocity to average speed is $1:1$.

(D) Average velocity is defined as the total displacement divided by the total time taken.
Displacement is the shortest distance between the initial point $A$ and the final point $B$.
Since the displacement is the same regardless of the path taken,the average velocity depends only on the total time taken to travel from $A$ to $B$.
Therefore,we can use any of the paths $1, 2$ or $3$ to calculate the average velocity,provided we know the total time taken for that specific path.
Thus,the correct answer is all of the above.

(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.

(A) $(i)$ Yes. $A$ particle moving with constant speed can be accelerated if its direction of motion changes,such as in uniform circular motion.
$(ii)$ No. Acceleration is defined as the rate of change of velocity. If velocity is constant,the rate of change of velocity is zero,meaning acceleration is zero.

(A) The speed of an object moving in a circular path is defined as the distance traveled per unit time.
For one complete round of a circular track with radius $r$,the distance traveled is equal to the circumference of the circle,which is $2 \pi r$.
Given that the time taken to complete this round is $t$,the speed $v$ is calculated as:
$v = \frac{\text{Distance}}{\text{Time}} = \frac{2 \pi r}{t}$.

(B) Distance is the total path length covered by an object,while displacement is the shortest distance between the initial and final positions.
For the magnitude of displacement to be equal to the distance,the object must move in a straight line in a single direction without any change in direction.
If the object changes its direction,the total path length (distance) will be greater than the straight-line distance (displacement) between the start and end points.

(A) Yes. When a body moves along a circular path with a constant speed,its direction of motion changes at every point. Since velocity is a vector quantity (having both magnitude and direction),a change in direction results in a change in velocity. This change in velocity constitutes acceleration,known as centripetal acceleration.

(C) Displacement is defined as the shortest distance between the initial position and the final position of an object.
In this scenario,both the boy and the football start from the same initial position (where the boy kicks the ball) and end at the same final position (where the boy catches the ball).
Since the initial and final positions are identical for both,their displacement is equal.
Although the path taken by the football is a curved trajectory (projectile motion) and the path taken by the boy is a straight or curved path on the ground,the displacement depends only on the start and end points,not the path taken.

(B) The distance covered by a body is equal to the magnitude of its displacement when the body moves along a straight-line path in a single direction.
In this scenario,the path length (distance) is identical to the straight-line distance between the initial and final positions (displacement).
If the body reverses its direction,the distance will be greater than the magnitude of the displacement.

(B) When a body moves with constant velocity,the displacement is directly proportional to time $(s = v \times t)$.
Since the relationship is linear,the displacement-time graph is a straight line.
If the body starts from the origin at time $t = 0$,the graph passes through the origin.

(B) In a displacement-time graph,the slope represents the velocity of the object.
When a body is in uniform motion,it covers equal displacements in equal intervals of time.
Therefore,the velocity remains constant.
Since the velocity is constant,the slope of the displacement-time graph is also constant.

(B) For a body moving with constant acceleration,the displacement $s$ is related to time $t$ by the kinematic equation $s = ut + \frac{1}{2}at^2$. Since $s$ is a quadratic function of $t$,the displacement-time graph is a parabola.

(B) No,it is not possible. Distance is the total path length covered by an object. If an object moves,it must cover some distance. Displacement is the shortest distance between the initial and final positions. If displacement is not zero,it means the object has changed its position,which implies that the object must have moved,and therefore,the distance travelled cannot be zero.

(B) When the magnitude of acceleration is constant but its direction is always perpendicular to the velocity of the particle,the particle undergoes uniform circular motion.
In this case,the acceleration acts as a centripetal acceleration,which changes the direction of the velocity vector continuously while keeping its magnitude constant.
Therefore,the path followed by the particle is a circular path.

(B) In a velocity$-$time graph,the distance travelled by an object is equal to the area enclosed between the velocity$-$time curve and the time axis.
For uniform motion,the velocity$-$time graph is a straight line parallel to the time axis.
The area of the rectangle formed under this line represents the distance covered by the body during a specific time interval.

(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.

(A) The displacement $s$ of a body is proportional to the time $t$ elapsed,which can be expressed as $s \propto t$ or $s = kt$,where $k$ is a constant representing velocity.
Since the velocity $(v = ds/dt = k)$ remains constant over time,the body is said to be in uniform motion.
Therefore,the correct answer is uniform motion.

(C) An object in motion is considered to be a point object if the distance it travels is very large compared to the dimensions (size) of the object itself. In such cases,the size of the object becomes negligible relative to the path covered.

(B) No,the speed of a body cannot be negative.
Speed is defined as the ratio of the total distance traveled to the time taken.
Since distance is a scalar quantity and represents the total path length covered,it is always positive or zero.
Therefore,the ratio of distance to time (speed) must also be non-negative.

(C) The velocity of a particle is defined as the rate of change of displacement with respect to time,which is given by the slope of the displacement-time graph $(v = \frac{ds}{dt})$.
If the displacement-time graph is parallel to the displacement axis,it implies that the displacement changes instantaneously at a single point in time.
Mathematically,the slope of a line parallel to the vertical axis is undefined or infinite.
Therefore,the velocity of the particle is infinite.