Textbook - MOTION Questions in English

(N/A) Yes,an object can have zero displacement even when it has moved through a distance.
This occurs when the final position of the object coincides with its initial position.
For example,if a person walks around a circular park and returns to the exact starting point,the total distance covered is equal to the circumference of the park,but the displacement is $0$ because the initial and final positions are the same.

(B) Given, side of the square field $= 10 \, m$.
Perimeter of the square field $= 4 \times 10 \, m = 40 \, m$.
The farmer completes one round of the field in $40 \, s$.
Total time taken $= 2 \, min \; 20 \, s = (2 \times 60) \, s + 20 \, s = 140 \, s$.
Number of rounds completed by the farmer $= \frac{\text{Total time}}{\text{Time for one round}} = \frac{140 \, s}{40 \, s} = 3.5 \, \text{rounds}$.
After $3$ complete rounds, the farmer will be back at the starting position. In the remaining $0.5$ round, the farmer will cover half the perimeter, i.e., $20 \, m$, and will reach the diagonally opposite corner of the square.
Let the starting point be $A$ and the diagonally opposite point be $C$. The displacement is the length of the diagonal $AC$.
Using the Pythagorean theorem, $AC = \sqrt{(10 \, m)^2 + (10 \, m)^2} = \sqrt{100 + 100} \, m = \sqrt{200} \, m = 10\sqrt{2} \, m$.
Taking $\sqrt{2} \approx 1.414$, the displacement $= 10 \times 1.414 \, m = 14.14 \, m$.

(D) Displacement is the shortest distance between the initial and final positions of an object.
$1$. Displacement can be zero if the initial and final positions of the object are the same (e.g., in circular motion).
$2$. The magnitude of displacement is always less than or equal to the distance travelled by the object $(|\text{Displacement}| \le \text{Distance})$.
Therefore, both statements $(a)$ and $(b)$ are false. The correct option is $(d)$.

(N/A)

Speed	Velocity
Speed is the distance travelled by an object in a given interval of time.	Velocity is the displacement of an object in a given interval of time.
$Speed = \frac{\text{distance}}{\text{time}}$	$Velocity = \frac{\text{displacement}}{\text{time}}$
Speed is a scalar quantity, i.e., it has only magnitude.	Velocity is a vector quantity, i.e., it has both magnitude and direction.

(B) The magnitude of average velocity is defined as the ratio of the total displacement to the total time taken. The average speed is defined as the ratio of the total distance traveled to the total time taken.
For an object moving in a straight line without changing its direction,the total distance traveled is equal to the magnitude of the total displacement.
Therefore,under the condition of motion in a straight line without changing direction,the magnitude of average velocity is equal to the average speed.

(B) The odometer is a device used in automobiles to measure the total distance traveled by the vehicle.
It records the cumulative distance covered from the time the vehicle was manufactured or since the last reset.

(B) Uniform motion is defined as the motion of an object in which it travels equal distances in equal intervals of time along a straight line.
Therefore,the path of an object in uniform motion is always a straight line.

(D) Given:
Speed of the signal $(v)$ = $3 \times 10^{8} \, m \, s^{-1}$
Time taken $(t)$ = $5 \, \text{minutes} = 5 \times 60 \, \text{seconds} = 300 \, s$
Formula:
Distance $(d)$ = $\text{Speed} \times \text{Time}$
Calculation:
$d = (3 \times 10^{8} \, m \, s^{-1}) \times (300 \, s)$
$d = 3 \times 10^{8} \times 3 \times 10^{2} \, m$
$d = 9 \times 10^{10} \, m$
Therefore, the distance of the spaceship from the ground station is $9 \times 10^{10} \, m$.

(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 $0.0185 \, m/s^{2}$.

(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.

If the distance-time graph is a straight line parallel to the time axis,it indicates that the distance of the object from the reference point remains constant as time passes.
This means the object is not changing its position with respect to time.
Therefore,the object is at rest.

(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.

(D) Given:
Initial velocity,$u = 0\, m s^{-1}$
Acceleration,$a = 0.1\, m s^{-2}$
Time,$t = 2\, min = 120\, s$
$(a)$ To find the speed acquired $(v)$:
Using the first equation of motion,$v = u + at$
$v = 0 + (0.1\, m s^{-2} \times 120\, s)$
$v = 12\, m s^{-1}$
$(b)$ To find the distance travelled $(s)$:
Using the second equation of motion,$s = ut + \frac{1}{2}at^2$
$s = (0 \times 120) + \frac{1}{2} \times 0.1 \times (120)^2$
$s = 0 + 0.05 \times 14400$
$s = 720\, m$
Thus,the speed acquired is $12\, m s^{-1}$ and the distance travelled is $720\, m$.

(A) Initial speed of the train,$u = 90\, km/h = 90 \times \frac{5}{18} = 25\, m/s$.
Final speed of the train,$v = 0\, m/s$ (as the train comes to rest).
Acceleration,$a = -0.5\, m/s^2$.
Using the third equation of motion,$v^2 = u^2 + 2as$.
Substituting the values: $(0)^2 = (25)^2 + 2(-0.5)s$.
$0 = 625 - 1s$.
$s = 625\, m$.
The train will cover a distance of $625\, m$ before it comes to rest.

(A) Initial velocity of the trolley,$u = 0 \,m \,s^{-1} = 0 \,cm \,s^{-1}$.
Acceleration,$a = 2 \,m \,s^{-2} = 200 \,cm \,s^{-2}$.
Time,$t = 3 \,s$.
Using the first equation of motion,$v = u + at$.
Substituting the values: $v = 0 + (200 \,cm \,s^{-2} \times 3 \,s)$.
$v = 600 \,cm \,s^{-1}$.
Therefore,the velocity of the trolley after $3 \,s$ is $600 \,cm \,s^{-1}$.

(C) Initial velocity of the car,$u = 0 \,m s^{-1}$.
Acceleration,$a = 4 \,m s^{-2}$.
Time,$t = 10 \,s$.
We use the second equation of motion: $s = ut + \frac{1}{2}at^2$.
Substituting the values: $s = (0 \times 10) + \frac{1}{2} \times 4 \times (10)^2$.
$s = 0 + \frac{1}{2} \times 4 \times 100$.
$s = 2 \times 100 = 200 \,m$.
Therefore,the distance covered by the car in $10 \,s$ is $200 \,m$.

(D) Given: Initial velocity $u = 5 \, m s^{-1}$.
Final velocity at maximum height $v = 0 \, m s^{-1}$.
Acceleration $a = -10 \, m s^{-2}$ (since it acts downwards).
Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - (5)^2 = 2 \times (-10) \times s$
$-25 = -20s$
$s = \frac{-25}{-20} = 1.25 \, m$.
Using the equation of motion $v = u + at$:
$0 = 5 + (-10)t$
$10t = 5$
$t = 0.5 \, s$.
Thus,the height attained is $1.25 \, m$ and the time taken is $0.5 \, s$.

(A) Diameter of circular track $(D) = 200\, m$.
Radius of circular track $(r) = D / 2 = 100\, m$.
Time taken for one round $(t) = 40\, s$.
Total time given $= 2\, min\, 20\, s = 140\, s$.
Number of rounds $= 140 / 40 = 3.5\, rounds$.
Distance covered $= 3.5 \times (2 \pi r) = 3.5 \times 2 \times (22/7) \times 100 = 7 \times (22/7) \times 100 = 2200\, m$.
After $3.5\, rounds$, the athlete will be at the diametrically opposite point of the starting position.
Displacement $= \text{Diameter of the circular track} = 200\, m$.

(B) The distance covered from $A$ to $B$ is $300 \, m$.
The time taken to travel from $A$ to $B$ is $2 \, min \, 30 \, sec = (2 \times 60) + 30 = 150 \, s$.
Average speed from $A$ to $B$ is defined as the total distance divided by the total time taken:
$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{300 \, m}{150 \, s} = 2 \, m s^{-1}$.
Velocity from $A$ to $B$ is defined as the displacement divided by the total time taken. Since the motion is in a straight line from $A$ to $B$,the displacement is equal to the distance $(300 \, m)$:
$\text{Velocity} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{300 \, m}{150 \, s} = 2 \, m s^{-1}$.
Thus,the average speed and velocity from $A$ to $B$ are both $2 \, m s^{-1}$.

(C) Let the distance between home and school be $S$.
Time taken to travel from home to school is $t_1 = S / 20$.
Time taken to travel from school to home is $t_2 = S / 40$.
Total distance covered = $S + S = 2S$.
Total time taken = $t_1 + t_2 = S / 20 + S / 40 = (2S + S) / 40 = 3S / 40$.
Average speed = $\text{Total distance} / \text{Total time} = 2S / (3S / 40) = (2S \times 40) / 3S = 80 / 3 = 26.67 \, km \, h^{-1}$.

(D) Given: Initial velocity of the motorboat,$u = 0 \, m s^{-1}$.
Acceleration of the motorboat,$a = 3.0 \, m s^{-2}$.
Time interval,$t = 8.0 \, s$.
We use the second equation of motion: $s = ut + \frac{1}{2}at^2$.
Substituting the given values:
$s = (0 \times 8.0) + \frac{1}{2} \times 3.0 \times (8.0)^2$.
$s = 0 + \frac{1}{2} \times 3.0 \times 64$.
$s = 3.0 \times 32$.
$s = 96 \, m$.
Therefore,the boat travels $96 \, m$ during this time.

(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 $9.14 \, km - 0 \, km = 9.14 \, km$.

(C) Let the final velocity with which the ball strikes the ground be $v$ and the time taken be $t$.
Initial velocity of the ball,$u = 0 \, m s^{-1}$.
Distance or height of fall,$s = 20 \, m$.
Downward acceleration,$a = 10 \, m s^{-2}$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
$v^2 = 2as + u^2$.
$v^2 = 2 \times 10 \times 20 + 0^2 = 400$.
Therefore,final velocity $v = \sqrt{400} = 20 \, m s^{-1}$.
Using the first equation of motion: $v = u + at$.
$20 = 0 + 10 \times t$.
Therefore,time $t = 20 / 10 = 2 \, s$.

(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.

(N/A) Possible.
When a ball is thrown vertically upwards,at its maximum height,its velocity is zero,but it still experiences a constant acceleration due to gravity $(g = 9.8\,m/s^2)$ acting downwards.
$(b)$ Possible.
When an object moves in a circular path with a uniform speed,its velocity changes continuously due to the change in direction. This change in velocity results in a centripetal acceleration directed towards the center of the circle.
$(c)$ Possible.
An object moving in a circular path is a classic example. The velocity of the object is tangential to the path,while the centripetal acceleration acts towards the center of the circle,which is perpendicular to the direction of motion.

(B) The radius of the circular orbit is $r = 42250 \, km = 42250000 \, m$.
The time taken to complete one revolution is $t = 24 \, h = 24 \times 60 \times 60 \, s = 86400 \, s$.
The speed $v$ of an object moving in a circular path is given by the formula $v = \frac{2 \pi r}{t}$.
Substituting the values: $v = \frac{2 \times (22/7) \times 42250000}{86400}$.
$v = \frac{2 \times 22 \times 42250000}{7 \times 86400} \approx 3073.74 \, m s^{-1}$.
Rounding to the nearest whole number,the speed is approximately $3074 \, m s^{-1}$.

(C) Total distance covered from $A$ to $C = AB + BC = 300 \, m + 100 \, m = 400 \, m$.
Total time taken from $A$ to $C = \text{Time for } AB + \text{Time for } BC = (2 \times 60 + 30) \, s + 60 \, s = 150 \, s + 60 \, s = 210 \, s$.
Average speed = $\text{Total distance} / \text{Total time} = 400 \, m / 210 \, s \approx 1.904 \, m s^{-1}$.
Displacement from $A$ to $C = AB - BC = 300 \, m - 100 \, m = 200 \, m$.
Average velocity = $\text{Displacement} / \text{Total time} = 200 \, m / 210 \, s \approx 0.952 \, m s^{-1}$.