Mix Example - MOTION Questions in English

(A) $(i)$ Graph $(a)$ represents this case. The ball starts with an initial velocity,experiences constant retardation due to gravity,and its velocity decreases to become zero at the highest point. Thereafter,the ball falls down,and its velocity increases in the opposite direction until it returns to the hand.
$(ii)$ Graph $(d)$ represents the case of a body decelerating (decreasing speed) to a constant speed and then accelerating (increasing speed).

(N/A) This graph represents the uniform acceleration of a body that starts from rest. At time $t = 0$,the velocity of the body is $0 \ m/s$.
$(b)$ This graph represents the uniform acceleration of a body that does not start from rest. At time $t = 0$,the body already possesses an initial velocity represented by the intercept on the velocity axis.

(N/A) Average velocity is defined as the ratio of total displacement to total time taken,whereas average speed is defined as the ratio of total distance travelled to total time taken.
Displacement is a vector quantity and can be zero if the initial and final positions of an object are the same (e.g.,in circular motion). Since average velocity depends on displacement,it can be zero.
Distance is a scalar quantity representing the total path length covered and is always positive for a moving body. Since average speed depends on total distance,it cannot be zero for a moving body.

(N/A) Velocity is determined by the slope of the displacement-time graph,where $V = \frac{\Delta \text{displacement}}{\Delta \text{time}}$.
$(i)$ For segment $A-B$: The displacement changes from $0 \ m$ to $3 \ m$ in the time interval from $2 \ s$ to $5 \ s$.
$V_{AB} = \frac{3 - 0}{5 - 2} = \frac{3}{3} = 1 \ m \ s^{-1}$.
$(ii)$ For segment $B-C$: The displacement remains constant at $3 \ m$ from $5 \ s$ to $7 \ s$.
$V_{BC} = \frac{3 - 3}{7 - 5} = \frac{0}{2} = 0 \ m \ s^{-1}$.
$(iii)$ For segment $C-D$: The displacement changes from $3 \ m$ to $0 \ m$ in the time interval from $7 \ s$ to $10 \ s$.
$V_{CD} = \frac{0 - 3}{10 - 7} = \frac{-3}{3} = -1 \ m \ s^{-1}$.

(N/A) $(i)$ The segment $OP$ indicates uniform speed.
Speed $v = \frac{\text{Distance}}{\text{Time}} = \frac{75 \ m - 0 \ m}{4 \ s - 0 \ s} = \frac{75}{4} = 18.75 \ m \ s^{-1}$.
(ii) The segment $PQ$ is parallel to the time axis,which indicates that the object is stationary. The duration is from $t = 4 \ s$ to $t = 14 \ s$,so the object was stationary for $14 - 4 = 10 \ s$.
(iii) The segment $QR$ represents a situation where the distance decreases as time increases,which is physically impossible for a single object moving in one dimension. Thus,this does not represent a real situation.

(N/A) Speed is the distance travelled per unit time,whereas velocity is the displacement per unit time.
Speed is a scalar quantity,while velocity is a vector quantity.
Speed can never be zero for a moving object,whereas velocity can be zero if the displacement is zero.
$(b)$ Both speed and velocity have the same $SI$ units,which is $m \ s^{-1}$.

(N/A) Uniform circular motion is defined as the motion of an object traveling along a circular path at a constant speed.
$(b)$ Yes,uniform circular motion is an accelerated motion. Even though the speed of the object remains constant,the direction of its velocity changes continuously at every point along the circular path. Since acceleration is defined as the rate of change of velocity,and velocity is a vector quantity (having both magnitude and direction),the continuous change in direction results in a continuous change in velocity,thereby causing acceleration.

$(a)$ This is a distance-time table.
$(b)$ The bus is in uniform motion. This conclusion is drawn because the bus covers an equal distance of $10\, km$ in equal intervals of time ($15$ minutes each).

(A) Free fall is defined as the motion of an object falling towards the Earth solely under the influence of gravitational force,with no other forces like air resistance acting upon it.
In a vacuum,there is no air resistance. The acceleration experienced by any object falling freely is the acceleration due to gravity $(g)$,which is approximately $9.8\,m/s^2$ near the Earth's surface.
Since the acceleration due to gravity $(g)$ is independent of the mass of the falling object,both bodies will experience the same acceleration.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (initial velocity) and $s = h$ (height),we get $h = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are the same for both bodies,the time taken $(t)$ will be identical for both. Therefore,both bodies will hit the ground at the same time.

(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) From the number line,the positions are: $A = 7 \, m$,$B = -2 \, m$,$C = 3 \, m$.
$(a)$ For motion from $A$ to $B$:
Total distance $= |(-2) - 7| = 9 \, m$.
Total displacement $= -2 - 7 = -9 \, m$.
Time taken $= 10 \, s$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{9}{10} = 0.9 \, m/s$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{-9}{10} = -0.9 \, m/s$.
$(b)$ For motion from $A$ to $C$ through $B$:
Total distance $= |(-2) - 7| + |3 - (-2)| = 9 + 5 = 14 \, m$.
Total displacement $= 3 - 7 = -4 \, m$.
Total time $= 10 + 5 = 15 \, s$.
Average speed $= \frac{14}{15} \approx 0.933 \, m/s$.
Average velocity $= \frac{-4}{15} \approx -0.267 \, m/s$.

(N/A) The velocity-time graph shows a constant velocity over time,which represents uniform motion.
$(b)$ The area under a velocity-time graph represents the displacement of the object. Therefore,the area of rectangle $OABC$ represents the displacement.
$(c)$ The straight line $AB$ is parallel to the time axis,which indicates that the velocity of the object remains constant over time,representing uniform motion.

(N/A) $(i)$ $A$ freely falling stone exhibits uniformly accelerated motion because it is under the influence of constant gravitational acceleration $(g \approx 9.8 \ m/s^2)$.
$(ii)$ When a stone is thrown vertically upwards,the earth exerts a gravitational force on it in the downward direction. This force acts against the direction of motion,causing retardation (negative acceleration),which leads to a continuous decrease in velocity.
$(iii)$ An example of such motion is an object moving in a circular path and returning to its starting point. Since the displacement is zero,the average velocity is zero,but the total distance covered is the circumference of the circle,so the average speed is not zero.

(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.

(N/A) To draw the distance-time graph, follow these steps:
$1$. Take the $x$-axis to represent time and the $y$-axis to represent distance in $km$.
$2$. Plot the given points: $(06:45, 0)$, $(07:00, 8)$, $(01:30, 8)$, and $(01:45, 0)$.
$3$. Join the points to obtain the graph. The graph shows that Renu travels $8 \, km$ to reach school, stays there for some time, and then returns home.
The distance-time graph is as shown in the provided image.

(N/A) Displacement is equal to the area under the velocity-time graph.
Displacement $= \text{Area of triangle} + \text{Area of rectangle} + \text{Area of triangle}$
$= (\frac{1}{2} \times 5 \times 4) + ((12 - 5) \times 4) + (\frac{1}{2} \times (15 - 12) \times 4)$
$= 10 + 28 + 6 = 44 \text{ m}$
$(b)$ The truck is decelerating when the velocity decreases with time,which occurs between $12 \text{ s}$ and $15 \text{ s}$.
$(c)$ Average velocity $= \frac{\text{Total displacement}}{\text{Total time}}$
$= \frac{44 \text{ m}}{15 \text{ s}} \approx 2.93 \text{ m s}^{-1}$

(A) When an object completes half of a circular path, the distance covered is equal to half of the circumference of the circle.
Distance $= \pi R = \frac{22}{7} \times 7 = 22 \, m$.
The displacement is the shortest distance between the initial and final positions, which is the diameter of the circle.
Displacement $= 2R = 2 \times 7 = 14 \, m$.
The object possesses uniform circular motion because it moves with a constant speed along a circular path.

(N/A) $(i)$ $A$ body is said to be in uniform motion if it covers equal distances in equal intervals of time.
$(ii)$ $A$ body is said to be in non-uniform motion if it covers unequal distances in equal intervals of time.
$(b)$ For an object in uniform motion,the distance-time graph is a straight line passing through the origin,as shown below:
[Graph: $A$ straight line starting from the origin $(0,0)$ on a Cartesian plane where the y-axis represents 'distance' and the x-axis represents 'time'.]

(N/A) The distance covered by an object is equal to the area under the speed-time graph.
For car $P$,the area is a triangle with base $= 4\, s$ and height $= 6\, m/s$.
Distance $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\, s \times 6\, m/s = 12\, m$.
For car $Q$,the area is a rectangle with length $= 4\, s$ and width $= 3\, m/s$.
Distance $= \text{length} \times \text{width} = 4\, s \times 3\, m/s = 12\, m$.
The difference in the distance travelled by the two cars is $12\, m - 12\, m = 0\, m$.
$(b)$ Yes,they move with the same speed at the point where the two graphs intersect. This occurs at $t = 2\, s$,where both cars have a speed of $3\, m/s$.
$(c)$ Car $P$ is undergoing uniform acceleration because its speed-time graph is a straight line passing through the origin,indicating a constant rate of change of speed. Car $Q$ is undergoing uniform motion (constant speed) because its speed-time graph is a horizontal line,indicating that its speed does not change with time.

(A) Given: Time $t = 3 \ s$,Final velocity $v = 0 \ m s^{-1}$ (at maximum height),Acceleration $a = g = -9.8 \ m s^{-2}$ (acting downwards).
To find initial velocity $u$:
Using the first equation of motion: $v = u + at$
$0 = u + (-9.8) \times 3$
$u = 29.4 \ m s^{-1}$.
To find maximum height $h$:
Using the second equation of motion: $h = ut + \frac{1}{2}at^2$
$h = (29.4 \times 3) + \frac{1}{2} \times (-9.8) \times (3)^2$
$h = 88.2 - 44.1$
$h = 44.1 \ m$.

(A) Part $1$: Acceleration phase
Initial velocity $u = 0 \, m/s$,Final velocity $v = 36 \, km/h = 36 \times (5/18) = 10 \, m/s$,Time $t_1 = 10 \, s$.
Acceleration $a_1 = (v - u) / t_1 = (10 - 0) / 10 = 1 \, m/s^2$.
Distance $s_1 = ut_1 + 0.5 a_1 t_1^2 = 0 + 0.5 \times 1 \times 10^2 = 50 \, m$.
Part $2$: Deceleration phase
Initial velocity $u = 10 \, m/s$,Final velocity $v = 0 \, m/s$,Time $t_2 = 20 \, s$.
Acceleration $a_2 = (v - u) / t_2 = (0 - 10) / 20 = -0.5 \, m/s^2$.
Distance $s_2 = ut_2 + 0.5 a_2 t_2^2 = 10 \times 20 + 0.5 \times (-0.5) \times 20^2 = 200 - 100 = 100 \, m$.
Total distance = $s_1 + s_2 = 50 + 100 = 150 \, m$.

(A) According to the law of conservation of momentum,the total momentum before the jump is equal to the total momentum after the jump.
Initially,the boat and the girl are moving together,so the initial momentum is $0$ (assuming the system starts from rest or considering the relative change).
Let $m_1 = 50\, kg$ be the mass of the girl and $v_1 = 3\, m s^{-1}$ be her velocity.
Let $m_2 = 300\, kg$ be the mass of the boat and $v_2$ be the velocity of the boat.
Using the principle of conservation of linear momentum: $m_1 v_1 + m_2 v_2 = 0$.
Substituting the values: $(50 \times 3) + (300 \times v_2) = 0$.
$150 + 300 v_2 = 0$.
$300 v_2 = -150$.
$v_2 = -\frac{150}{300} = -0.5\, m s^{-1}$.
The negative sign indicates that the boat moves in the opposite direction (backwards) with a velocity of $0.5\, m s^{-1}$.

(N/A) Average velocity is defined as the total displacement divided by the total time taken. Since displacement is a vector quantity,if an object returns to its starting point,the total displacement becomes $0$,resulting in an average velocity of $0$.
Average speed is defined as the total distance traveled divided by the total time taken. Since distance is a scalar quantity and represents the total path length,it cannot be $0$ for a moving object.
Example: Consider an object moving along a circular path of radius $r$. If the object completes one full revolution,the total displacement is $0$,so the average velocity is $0$. However,the total distance traveled is equal to the circumference of the circle,which is $2 \pi r$. Thus,the average speed is $\frac{2 \pi r}{t}$,where $t$ is the time taken.

(N/A) To calculate the total distance and displacement,we represent the motion on a coordinate system.
$1$. Total Distance:
Distance is the total path length covered by the hiker.
Total Distance $= 700\, m + 300\, m + 400\, m + 600\, m + 1200\, m + 300\, m + 100\, m = 3600\, m$.
$2$. Displacement:
Displacement is the shortest distance between the initial and final positions.
Let the starting point $O$ be $(0, 0)$.
- Move $700\, m$ North: Position $(0, 700)$.
- Move $300\, m$ East: Position $(300, 700)$.
- Move $400\, m$ North: Position $(300, 1100)$.
- Move $600\, m$ West: Position $(300 - 600, 1100) = (-300, 1100)$.
- Move $1200\, m$ South: Position $(-300, 1100 - 1200) = (-300, -100)$.
- Move $300\, m$ East: Position $(-300 + 300, -100) = (0, -100)$.
- Move $100\, m$ North: Position $(0, -100 + 100) = (0, 0)$.
Since the final position $(0, 0)$ is the same as the initial position,the displacement is $0\, m$.

(N/A) $(i)$ Taking house $A$ as the origin $(0 \ m)$,the positions of the houses are: $A = 0 \ m, B = 10 \ m, C = 20 \ m, D = 30 \ m, E = 40 \ m$.
$(ii)$ Position of house $C = 20 \ m$ and position of house $E = 40 \ m$. The position of $C$ relative to $E$ is $20 \ m - 40 \ m = -20 \ m$. This means house $C$ is $20 \ m$ to the left of house $E$.
$(iii)$ Taking house $B$ as the reference point $(0 \ m)$,the position of house $A$ is $-10 \ m$ and the position of house $D$ is $20 \ m$.

(N/A) $(i)$ Person $B$ is moving with constant speed because he travels an equal distance of $5\, km$ in equal intervals of time (every $5\, minutes$).
$(ii)$ Person $C$ has travelled the maximum distance, which is $10\, km$ between $8.00\, am$ and $8.05\, am$.
$(iii)$ Average speed of person $A = \frac{\text{Total distance travelled by } A}{\text{Total time taken}}$
$= \frac{25\, km}{20\, min} = \frac{25\, km}{(20/60)\, h} = \frac{25 \times 60}{20}\, km\, h^{-1} = 75\, km\, h^{-1}$.

(N/A) $(i)$ The motion of a car when it starts from rest and increases its velocity at a constant rate in a particular interval of time.
$(ii)$ The motion of a train when it slows down on approaching a station.
$(iii)$ The motion of a ball falling down freely.

(N/A) $(i)$ The slope of the distance$-$time graph represents the speed of the object.
$(ii)$ The area under a velocity$-$time graph represents the total displacement or distance covered by the object.
$(iii)$ The slope of a velocity$-$time graph represents the acceleration of the object.

(N/A) $(i)$ $A$ horizontal line indicates that the position of the object does not change with time,meaning the object is at rest.
$(ii)$ $A$ straight diagonal line indicates a constant slope,which means the object is moving with uniform velocity (constant speed).
$(iii)$ $A$ curved line indicates a changing slope,which means the object is moving with non-uniform velocity (variable speed/acceleration).

(N/A) $(i)$ At $t = 2 \, s$,position $x = 0 \, m$. At $t = 6 \, s$,position $x = 5 \, m$. Therefore,distance traveled $= 5 \, m - 0 \, m = 5 \, m$.
(ii) The speed is represented by the slope of the position-time graph. The slope is steeper between $2 \, s$ and $4 \, s$ compared to the interval between $4 \, s$ and $6 \, s$. Thus,the speed was greater during the interval $2 \, s$ to $4 \, s$.
(iii) Average speed $= \frac{\text{Total distance covered}}{\text{Total time taken}}$.
Total distance $= 5 \, m$.
Total time $= 6 \, s - 2 \, s = 4 \, s$.
Average speed $= \frac{5 \, m}{4 \, s} = 1.25 \, m/s$.

(N/A) Consider the velocity-time graph shown in the figure.
The distance $S$ covered by an object moving with uniform acceleration $a$ is equal to the area under the velocity-time graph,which is the area of the trapezium $OABD$.
Area of trapezium $OABD = \frac{(OA + BD) \times OD}{2}$
From the graph,we have:
$OA = u$ (initial velocity)
$BD = v$ (final velocity)
$OD = t$ (time taken)
Substituting these values,we get:
$S = \frac{(u + v) t}{2}$ --- (Equation $1$)
We know that for uniform acceleration,$v = u + at$,which can be rearranged to find time $t$:
$at = v - u$
$t = \frac{v - u}{a}$ --- (Equation $2$)
Substituting the value of $t$ from Equation $2$ into Equation $1$:
$S = \frac{(u + v)(v - u)}{2a}$
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we get:
$S = \frac{v^2 - u^2}{2a}$
Rearranging the terms,we get the final equation:
$v^2 - u^2 = 2aS$

(N/A) Distance is the total path length covered by an object,whereas displacement is the shortest distance between the initial and final positions of an object.
Distance is a scalar quantity,while displacement is a vector quantity.
Distance can never be negative or zero for a moving object,whereas displacement can be positive,negative,or zero.
$(b)$ The magnitude of average velocity is equal to the average speed when an object moves along a straight-line path in a single direction without changing its direction of motion.

(N/A) Acceleration is defined as the rate of change of velocity of an object with respect to time. Mathematically,$a = \frac{v - u}{t}$. Its $SI$ unit is $m s^{-2}$.
$(b)$ The velocity-time graphs are as follows:
$(i)$ For uniformly accelerated velocity,the graph is a straight line sloping upwards from the origin (or starting from a non-zero initial velocity),indicating that velocity increases at a constant rate over time.
$(ii)$ For uniformly retarded velocity,the graph is a straight line sloping downwards,indicating that velocity decreases at a constant rate over time until it reaches zero.

(N/A) The distance travelled by the car in the first $2 \ s$ is equal to the area under the velocity-time graph from $t = 0$ to $t = 2 \ s$,which is the area of $\Delta ABE$.
Distance $= \text{Area of } \Delta ABE = \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times 2 \ s \times 15 \ m/s = 15 \ m$.
$(ii)$ To find the braking force,we first calculate the acceleration (deceleration) from $t = 5 \ s$ to $t = 6 \ s$ using the slope of the line $CD$.
Initial velocity at $t = 5 \ s$ is $u = 15 \ m/s$.
Final velocity at $t = 6 \ s$ is $v = 0 \ m/s$.
Time interval $\Delta t = 6 \ s - 5 \ s = 1 \ s$.
Acceleration $a = \frac{v - u}{\Delta t} = \frac{0 - 15 \ m/s}{1 \ s} = -15 \ m/s^2$.
Using Newton's second law of motion,$F = m \times a$.
$F = 1000 \ kg \times (-15 \ m/s^2) = -15000 \ N$.
The negative sign indicates that the force is a braking force acting opposite to the direction of motion. Thus,the magnitude of the braking force is $15000 \ N$.

(N/A) For upward motion:
$v = u - gt$
At the highest point,final velocity $v = 0$.
So,$0 = u - gt_1$,which gives $t_1 = \frac{u}{g}$ $....(1)$
For downward motion:
The body starts from rest at the highest point,so initial velocity $u' = 0$.
Using $s = ut + \frac{1}{2}at^2$,where $s = h$ (maximum height) and $a = g$:
$h = 0 + \frac{1}{2}gt_2^2 \implies t_2 = \sqrt{\frac{2h}{g}}$.
Since $h = \frac{u^2}{2g}$,substituting this gives $t_2 = \sqrt{\frac{2(u^2/2g)}{g}} = \sqrt{\frac{u^2}{g^2}} = \frac{u}{g}$ $....(2)$
Comparing $(1)$ and $(2)$,$t_1 = t_2$. Thus,the time of ascent is equal to the time of descent.

(N/A) Since the body travels a distance $h$ upwards and $h$ downwards,the total distance travelled is $h + h = 2h$.
$(b)$ As the body returns to its starting point,the displacement is $0$.
The magnitude of the displacement is equal to the distance travelled if the object moves along a straight line in a single direction without changing its direction.

(N/A) When a body is stationary,its distance from the origin does not change with time. The graph is a straight line parallel to the time axis.
$(b)$ When a body is moving with a uniform speed,it covers equal distances in equal intervals of time. The graph is a straight line passing through the origin.
$(c)$ When a body is moving with non$-$uniform speed,it covers unequal distances in equal intervals of time. The graph is a curved line.

$(6.67\, m s^{-1})$ Average speed of an object is defined as the total distance travelled divided by the total time taken.
Step $1$: Calculate the total distance covered.
Distance for the first part $= 10\, m s^{-1} \times 5\, s = 50\, m$.
Distance for the second part $= 5\, m s^{-1} \times 10\, s = 50\, m$.
Total distance $= 50\, m + 50\, m = 100\, m$.
Step $2$: Calculate the total time taken.
Total time $= 5\, s + 10\, s = 15\, s$.
Step $3$: Calculate the average speed.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{100\, m}{15\, s} \approx 6.67\, m s^{-1}$.

(B) The odometer of an automobile measures the distance traveled by the vehicle.
To compare the speeds,we convert both into $m \, s^{-1}$:
$(i)$ Speed of scooter $= 300 \, m / 60 \, s = 5 \, m \, s^{-1}$.
$(ii)$ Speed of car $= 36 \, km/h = 36 \times (1000 \, m / 3600 \, s) = 36 \times (5/18) \, m \, s^{-1} = 10 \, m \, s^{-1}$.
Since $10 \, m \, s^{-1} > 5 \, m \, s^{-1}$,the car is moving faster than the scooter.

(N/A) If the velocity-time graph is a straight line parallel to the time axis,the object moves with a constant velocity and the motion is uniform.
$(b)$ Since velocity is constant,the change in velocity is zero. Therefore,acceleration,which is the rate of change of velocity,is zero.
$(c)$ For uniform motion,the distance-time graph is a straight line passing through the origin,as shown in the image.

(N/A) The velocity-time graph for a body moving with constant acceleration $a$ and initial velocity $u$ is shown in the figure.
The distance covered $S$ by the object in time $t$ is equal to the area under the velocity-time graph.
The area under the graph is the sum of the area of the rectangle $OACD$ and the area of the triangle $ABC$ on top of it.
The area of the rectangle $OACD$ is given by: $\text{Area} = \text{length} \times \text{breadth} = t \times u = ut$.
The area of the triangle $ABC$ is given by: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times (v - u)$.
Since the total distance $S$ is the sum of these areas,we have:
$S = ut + \frac{1}{2} \times t \times (v - u)$ $....(1)$
From the first equation of motion,we know that $v = u + at$,which implies $v - u = at$.
Substituting $(v - u) = at$ into equation $(1)$,we get:
$S = ut + \frac{1}{2} \times t \times (at)$
$S = ut + \frac{1}{2}at^{2}$

(N/A) $(i)$ The part $AB$ of the graph shows acceleration,as the velocity increases with time.
$(ii)$ The part $CD$ of the graph shows retardation (deceleration),as the velocity decreases with time.
$(iii)$ The distance travelled is equal to the area under the $v-t$ graph.
For the first $4 \ s$,the area is represented by the triangle $ABE$.
Distance $S = \text{Area of } \Delta ABE = \frac{1}{2} \times \text{base} \times \text{height}$
$S = \frac{1}{2} \times (AE) \times (BE)$
$S = \frac{1}{2} \times 4 \ s \times 4 \ m/s = 8 \ m$.

(N/A) Distance: It is defined as the total path length covered by a body.
Displacement: It is defined as the shortest distance between the initial and final positions of a body.
Given that the circumference of the circular park is $176 \, m$. The body completes one revolution in $4$ minutes.
After $6$ minutes,the number of revolutions completed $= \frac{6}{4} = 1.5$ revolutions.
Since $1$ revolution brings the body back to the starting point,after $1.5$ revolutions,the body will be at a position diametrically opposite to the starting point.
Given $2 \pi r = 176 \, m$,where $r$ is the radius.
$r = \frac{176}{2 \pi} = \frac{176 \times 7}{2 \times 22} = 28 \, m$.
The displacement after $1.5$ revolutions is the diameter of the circular path.
Displacement $= 2r = 2 \times 28 = 56 \, m$.

(N/A) Displacement is equal to the area under the velocity$-$time graph.
The graph consists of a triangle from $t = 0$ to $t = 5$ s, a rectangle from $t = 5$ to $t = 12$ s, and a triangle from $t = 12$ to $t = 15$ s.
Area $= (1/2 \times \text{base} \times \text{height}) + (\text{length} \times \text{breadth}) + (1/2 \times \text{base} \times \text{height})$
Area $= (1/2 \times 5 \times 4) + (7 \times 4) + (1/2 \times 3 \times 4) = 10 + 28 + 6 = 44 \text{ m}$.
$(b)$ Deceleration occurs when the velocity decreases with time. This corresponds to the part of the graph where the slope is negative, which is from $t = 12$ s to $t = 15$ s.
$(c)$ Average velocity $= \text{Total displacement} / \text{Total time} = 44 \text{ m} / 15 \text{ s} = 2.93 \text{ m s}^{-1}$.

(N/A) At $t=0$,the speed is $0$ and the slope of the graph is $0$. Therefore,the initial acceleration is $0 \, m/s^2$.
$(b)$ Between $t=5 \, s$ and $t=60 \, s$,the slope of the speed-time graph is increasing. Since the slope of a speed-time graph represents acceleration,the acceleration of the rocket is increasing during this interval.
$(c)$ At $t=80 \, s$,the graph is a straight line. The slope of this line is constant. Calculating the slope between $t=60 \, s$ $(v \approx 480 \, m/s)$ and $t=100 \, s$ $(v = 1400 \, m/s)$: $a = \frac{1400 - 480}{100 - 60} = \frac{920}{40} = 23 \, m/s^2$. The acceleration is constant because the graph is linear in this region.
$(d)$ Using Newton's second law,$F = ma$. Given $m = 1.6 \times 10^{6} \, kg$ and $a = 23 \, m/s^2$,the resultant force $F = 1.6 \times 10^{6} \times 23 = 3.68 \times 10^{7} \, N$. The force is non-zero because the rocket is accelerating.

(N/A) Consider a velocity$-$time graph where an object starts with initial velocity $u$ at time $t=0$ and reaches final velocity $v$ at time $t$. The slope of the velocity$-$time graph represents acceleration $(a)$.
Slope $= \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{v - u}{t}$.
Therefore,$a = \frac{v - u}{t}$,which gives $at = v - u$,or $v = u + at$.
$(b)$ The device used to measure the distance travelled by a vehicle is called an odometer.
$(c)$ Yes,the displacement of a moving object can be zero. This happens if the object returns to its initial starting position after completing its motion,as displacement is the shortest distance between the initial and final positions.

$(a)$ Acceleration is the slope of the velocity-time graph.
For car $B$, between $t = 2 \, s$ and $t = 4 \, s$, the velocity changes from $20 \, m/s$ to $40 \, m/s$.
$a = \frac{v_2 - v_1}{t_2 - t_1} = \frac{40 - 20}{4 - 2} = \frac{20}{2} = 10 \, m/s^2$.
$(b)$ Both cars have the same velocity where their graphs intersect. Looking at the graph, the intersection point is at $t = 2 \, s$ (velocity $= 20 \, m/s$) and $t = 6 \, s$ (velocity $= 60 \, m/s$).
$(c)$ Distance travelled is the area under the velocity-time graph.
For car $A$: The area is a trapezoid from $t = 1$ to $t = 4$ plus a rectangle from $t = 4$ to $t = 8$.
Area $= (\frac{1}{2} \times (4 - 1) \times 60) + (60 \times (8 - 4)) = (\frac{1}{2} \times 3 \times 60) + (60 \times 4) = 90 + 240 = 330 \, m$.
For car $B$: The area is a triangle from $t = 0$ to $t = 8$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 80 = 320 \, m$.
$(d)$ After $8 \, s$, car $A$ has travelled $330 \, m$ and car $B$ has travelled $320 \, m$.
Therefore, car $A$ is ahead by $330 - 320 = 10 \, m$.

(N/A) $(i)$ Looking at the graph,the girl reaches the $50 \ m$ mark (friend's house) at $14$ minutes.
$(ii)$ From $0$ to $2$ min,distance $= 20 \ m$. From $2$ to $4$ min,distance $= 0 \ m$. From $4$ to $6$ min,distance $= 40 - 20 = 20 \ m$. From $6$ to $8$ min,distance $= 0 \ m$. From $8$ to $10$ min,distance $= 40 - 20 = 20 \ m$. From $10$ to $12$ min,distance $= 0 \ m$. Total distance $= 20 + 0 + 20 + 0 + 20 + 0 = 60 \ m$.
$(iii)$ She moves towards her house when the distance from the house decreases. This happens during the intervals $8$ to $10$ minutes and $14$ to $16$ minutes.
$(iv)$ She is at rest when the distance remains constant (horizontal line). This occurs during $2-4$ min ($2$ min),$6-8$ min ($2$ min),and $10-12$ min ($2$ min). Total time at rest $= 2 + 2 + 2 = 6$ minutes.
$(v)$ She returns home from $14$ to $16$ minutes. Distance covered $= 50 \ m$. Time taken $= 2$ minutes. Speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{50 \ m}{2 \ min} = 25 \ m/min$.

(N/A) $(i)$ At $t = 0 \ h$,bus $B$ is at $20 \ km$ and bus $A$ is at $0 \ km$. Thus,bus $B$ was ahead by $20 \ km$.
$(ii)$ Yes,they meet at the point where the two lines intersect,which is at $t = 9 \ h$.
$(iii)$ At the point of intersection $(t = 9 \ h)$,the distance on the $y$-axis is $60 \ km$. Thus,bus $A$ has travelled $60 \ km$.
$(iv)$ At $t = 12 \ h$,bus $A$ is at $80 \ km$ and bus $B$ is at $70 \ km$. The difference is $80 \ km - 70 \ km = 10 \ km$.
$(v)$ Bus $A$ is moving faster because the slope of its distance-time graph is steeper than that of bus $B$,indicating a higher speed.

(N/A) $(ii)$ According to the velocity-time graph:
Distance moved by the body after $2\, s$ is equal to the area under the graph from $t = 0$ to $t = 2\, s$.
Area $= 2\, m s^{-1} \times 2\, s = 4\, m$.
Distance covered by the body after $12\, s$ is the total area under the graph from $t = 0$ to $t = 12\, s$.
This area consists of the rectangle for the first $5\, s$,the trapezoid for the next $5\, s$,and the trapezoid for the final $2\, s$ (from $t = 10\, s$ to $t = 12\, s$).
Total Distance $= (2\, m s^{-1} \times 5\, s) + \frac{1}{2} \times (2\, m s^{-1} + 10\, m s^{-1}) \times 5\, s + \frac{1}{2} \times (10\, m s^{-1} + 6\, m s^{-1}) \times 2\, s$
$= 10\, m + 30\, m + 16\, m = 56\, m$.