Graphical Representation of Motion Questions in English

(N/A) $1$. For uniform motion: The distance-time graph is a straight line. This indicates that the object covers equal distances in equal intervals of time,meaning the speed is constant.
$2$. For non-uniform motion: The distance-time graph is not a straight line; it is a curve. This indicates that the object covers unequal distances in equal intervals of time,meaning the speed is changing.

(N/A) If the speed-time graph is a straight line parallel to the time axis,it indicates that the speed of the object remains constant over time.
This type of motion is known as uniform motion,where the acceleration of the object is zero.

(A) The area under the velocity-time graph represents the displacement or the distance covered by an object moving with a certain velocity over a specific time interval.

(A) First,convert the initial speeds from $km\, h^{-1}$ to $m\, s^{-1}$:
For the first car: $u_1 = 52 \times (1000 / 3600) \, m\, s^{-1} = 14.44 \, m\, s^{-1}$.
For the second car: $u_2 = 3 \times (1000 / 3600) \, m\, s^{-1} = 0.83 \, m\, s^{-1}$.
The distance travelled by a car is equal to the area under the speed-time graph.
Distance travelled by the first car = Area of triangle with base $5\, s$ and height $14.44 \, m\, s^{-1}$.
$= (1 / 2) \times 5 \times 14.44 = 36.1 \, m$.
Distance travelled by the second car = Area of triangle with base $10\, s$ and height $0.83 \, m\, s^{-1}$.
$= (1 / 2) \times 10 \times 0.83 = 4.15 \, m$.
Comparing the two distances,the first car travelled farther after the brakes were applied.

(B) Object $B$.
$(b)$ No,all three objects $A$,$B$,and $C$ never meet at a single point on the graph.
$(c)$ To find the distance $C$ has travelled when $B$ passes $A$:
$1$. Identify the point where $B$ and $A$ intersect. This occurs at a distance of $9 \, km$ from the origin.
$2$. At this time,look at the position of object $C$ on the graph. Object $C$ is at a distance of approximately $7 \, km$ from the origin.
$3$. Object $C$ started at $2 \, km$ from the origin.
$4$. Therefore,the distance travelled by $C$ is $7 \, km - 2 \, km = 5 \, km$.
$(d)$ To find the distance $B$ has travelled by the time it passes $C$:
$1$. Identify the point where $B$ and $C$ intersect. This occurs at a distance of approximately $9.14 \, km$ from the origin.
$2$. Since object $B$ started from the origin $(0 \, km)$,the distance travelled by $B$ is $9.14 \, km - 0 \, km = 9.14 \, km$.

(N/A) The distance travelled by the car is equal to the area under the speed-time graph. To find the distance travelled in the first $4 \, s$,we calculate the area under the curve from $t = 0 \, s$ to $t = 4 \, s$. By observing the grid,the area is approximately a triangle with base $4 \, s$ and height $6 \, m/s$. Thus,the distance is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \, s \times 6 \, m/s = 12 \, m$.
$(b)$ Uniform motion is represented by a horizontal line in a speed-time graph,where the speed remains constant over time. In this graph,the part from $6 \, s$ to $10 \, s$ is a horizontal line,which represents the uniform motion of the car.

(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.

(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.

(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) $(i)$ 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.

(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 $v = \frac{\Delta x}{\Delta t} = \frac{0 - 6}{16 - 10} = \frac{-6}{6} = -1 \, m/s$.

(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) 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) 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.

(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.

(C) The velocity of a particle is defined as the rate of change of displacement with respect to time, given by the slope of the displacement $-$ time graph.
If the displacement $-$ time graph is parallel to the time axis, it means that the displacement of the particle does not change as time passes.
Since the displacement remains constant, the change in displacement $(\Delta s)$ is $0$.
Therefore, velocity $(v = \frac{\Delta s}{\Delta t})$ is $0$.

The graph shows a straight line passing through the origin,representing a distance-time graph where the distance covered is directly proportional to the time taken.
This indicates that the object is moving with a constant speed.
Therefore,the type of motion represented is uniform motion.

(B) In a displacement-time graph,the slope of the line represents the velocity of the object.
Since the slope of the line for car $B$ is steeper than the slope of the line for car $A$,the velocity of car $B$ is greater than that of car $A$.

(C) In a displacement$-$time graph,the slope represents the velocity of the body.
If the graph is a straight line parallel to the time axis,the displacement of the body does not change with time.
This means the velocity of the body is $0 \ m/s$.
Therefore,the body is at rest.

(N/A) The graph shows a horizontal line parallel to the time axis,which indicates that the distance of the object does not change with time. Therefore,the object is at rest.
$(b)$ The graph shows a straight line for some time (uniform motion) followed by a horizontal line (rest). Since the speed is not constant throughout the entire duration,this represents non-uniform motion.

(N/A) Similarity: Both graphs are velocity-time graphs.
Dissimilarity: In the first graph,the body starts its journey from rest (initial velocity $u = 0$) with uniform acceleration. In the second graph,the body starts its journey with a specific initial velocity $(u > 0)$ and increases its velocity with uniform acceleration.

(N/A) $(i)$ The graph shows a straight line with a negative slope,which indicates uniform retardation (deceleration) of the object.
$(ii)$ Acceleration $(a)$ is given by the slope of the velocity-time graph.
$a = \text{Slope} = \frac{v_2 - v_1}{t_2 - t_1}$
From the graph,at $t_1 = 0 \text{ s}$,$v_1 = 50 \text{ m s}^{-1}$ and at $t_2 = 40 \text{ s}$,$v_2 = 0 \text{ m s}^{-1}$.
$a = \frac{0 - 50 \text{ m s}^{-1}}{40 - 0 \text{ s}} = \frac{-50}{40} \text{ m s}^{-2} = -1.25 \text{ m s}^{-2}$

(N/A) $(i)$ When a body moves with uniform velocity,the distance covered is directly proportional to the time taken $(s \propto t)$. This results in a linear graph.
$(ii)$ When a body moves with variable velocity,the distance covered is not directly proportional to the time taken. The rate of change of distance with respect to time is non-constant,resulting in a non-linear graph.

(N/A) The velocity-time graphs have the following uses:
$(i)$ To determine the total distance travelled by a particle,which is equal to the area under the velocity-time graph.
$(ii)$ To determine the instantaneous velocity of the particle at any given time $t$.
$(iii)$ To determine the acceleration of the object,which is equal to the slope of the velocity-time graph.

(N/A) $(i)$ By finding the slope of the displacement-time graph, where $v = \frac{\Delta \text{displacement}}{\Delta \text{time}}$.
$(ii)$ By finding the slope of the velocity-time graph, where $a = \frac{\Delta \text{velocity}}{\Delta \text{time}}$.
$(iii)$ By finding the area under the velocity-time graph, where $\text{Displacement} = \text{Area under } v-t \text{ graph}$.
$(iv)$ By finding the area under the acceleration-time graph, where $\Delta v = \text{Area under } a-t \text{ graph}$.

(N/A) The slope of a displacement$-$time graph represents the velocity of the object.
No,a displacement$-$time graph cannot be parallel to the displacement axis.
This is because a line parallel to the displacement axis would imply that the object covers an infinite displacement in zero time,which is physically impossible.

(N/A) $(i)$ When a body is stationary,its displacement does not change with time. Therefore,the displacement-time graph is a straight line parallel to the time axis.
$(ii)$ When a body is moving with uniform velocity,it covers equal displacements in equal intervals of time. Therefore,the displacement-time graph is a straight line inclined to the time axis.
$(iii)$ When a body is moving with variable velocity,it covers unequal displacements in equal intervals of time. Therefore,the displacement-time graph is a curved line.

(N/A) The velocity-time graphs for the given situations are as follows:
$(i)$ For uniform velocity: The graph is a straight line parallel to the time axis,indicating that velocity does not change with time.
$(ii)$ For variable velocity with uniform acceleration: The graph is a straight line inclined to the time axis with a positive slope,indicating a constant rate of increase in velocity.
$(iii)$ For variable velocity with uniform retardation: The graph is a straight line inclined to the time axis with a negative slope,indicating a constant rate of decrease in velocity.
$(iv)$ For variable velocity and variable acceleration: The graph is a curve,indicating that the rate of change of velocity (acceleration) is not constant.

(N/A) $(i)$ The graph is a horizontal line parallel to the time axis,which indicates that the distance of the body from the origin does not change with time. Therefore,the body is stationary.
$(ii)$ The graph is a straight line passing through the origin,which indicates that the body covers equal distances in equal intervals of time. Therefore,the body is moving with uniform velocity (or uniform speed).
$(iii)$ The graph shows that initially,the distance increases linearly with time,indicating uniform velocity. After a certain point,the graph becomes a horizontal line,meaning the distance remains constant over time. Therefore,the body initially moves with uniform velocity and then stops suddenly.

(N/A) $(i)$ $A$ car moving with uniform velocity on a straight road.
$(ii)$ $A$ ball dropped from a height,falling freely under the action of gravity.
$(iii)$ $A$ train starting from a station,accelerating to a constant speed,moving with uniform velocity for some time,and then applying brakes to come to rest.

(N/A) $(i)$ Uniform motion: The velocity of the body remains constant over time,meaning the acceleration is zero.
$(ii)$ Uniformly accelerated motion: The velocity of the body increases linearly with time,indicating a constant positive acceleration.
$(iii)$ Uniformly retarded motion (or deceleration): The velocity of the body decreases linearly with time,indicating a constant negative acceleration.

$(D)$ Graph $D$ is not possible. This graph shows that at a given instant of time, the particle has two velocities. Also, it shows that at some point, it has infinite acceleration (graph parallel to the velocity axis). Both these conditions cannot be achieved in practice; therefore, this graph is also not possible.
$(b)$ Ratan exhibits values such as concern for his peers, helpfulness, and intelligence.

(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) 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.

$(a)$ 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).

(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) 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 $= \frac{\text{Total displacement}}{\text{Total time}}$
$= \frac{44 \text{ m}}{15 \text{ s}} \approx 2.93 \text{ m s}^{-1}$

(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) $(i)$ 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) 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.

(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.