Speed and Velocity Questions in English

(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.

(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 $9 \times 10^{10} \, m$.

(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) 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) 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 $1:1$.

(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.

(N/A) Average velocity is defined as the total displacement divided by the total time taken. When a particle returns to its starting point,the total displacement is $0$. Therefore,the average velocity is $0/t = 0$.
Average speed is defined as the total distance traveled divided by the total time taken. Since distance is a scalar quantity and is always positive for a moving object,the total distance covered cannot be $0$ if the particle has moved. Thus,the average speed cannot be zero.

(A) For an object moving with uniform velocity,the distance travelled is directly proportional to the time elapsed.
This means that the object covers equal distances in equal intervals of time,no matter how small these intervals are.
Mathematically,this is expressed as $s = v \times t$,where $s$ is distance,$v$ is uniform velocity,and $t$ is time.

(A) The direction of motion of an object is determined by its velocity. Velocity is a vector quantity that includes both the magnitude (speed) and the direction of motion. Acceleration,on the other hand,represents the rate of change of velocity. While acceleration can change the magnitude or the direction of the velocity,it does not define the current direction of motion itself.

(N/A) When an object is thrown vertically upwards, its velocity is directed upwards because it is moving away from the ground. However, the acceleration due to gravity $(g)$ acts downwards towards the center of the Earth. Since the object continues to move upwards despite the downward acceleration, it proves that the direction of motion is determined by the velocity and not by the acceleration.

(N/A) Velocity is defined as the rate of change of displacement of an object. It is a vector quantity,meaning it has both magnitude and direction. The mathematical formula for velocity $(v)$ is given by $v = \frac{\Delta s}{\Delta t}$,where $\Delta s$ is the change in displacement and $\Delta t$ is the time interval. Its $SI$ unit is meters per second $(m/s)$.

(N/A) Speed is defined as the rate of change of distance with respect to time. It is a scalar quantity,meaning it has only magnitude and no direction. It is always positive.
Velocity is defined as the rate of change of displacement with respect to time. It is a vector quantity,meaning it has both magnitude and direction. It can be positive,negative,or zero depending on the direction of motion.

(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.

$(6.67\, m s^{-1})$ 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 $= \frac{\text{Total distance}}{\text{Total time}} = \frac{100\, m}{15\, s} \approx 6.67\, m s^{-1}$.

(N/A) Differences between speed and velocity:
$1$. Speed is the distance traveled per unit time,whereas velocity is the displacement per unit time.
$2$. Speed is a scalar quantity (has only magnitude),whereas velocity is a vector quantity (has both magnitude and direction).
$(a)$ $A$ body is said to have uniform velocity if it covers equal displacements in equal intervals of time in a fixed direction.
$(b)$ $A$ body is said to have variable velocity if it covers unequal displacements in equal intervals of time or if its direction of motion changes.
When the velocity of a body changes at a non-uniform rate,the average velocity is calculated by taking the arithmetic mean of the initial velocity $(u)$ and the final velocity $(v)$:
$\text{Average Velocity} = \frac{u + v}{2}$

(N/A) Average speed is given by the formula:
$V_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$
The distance travelled by the vehicle is:
$1096 \text{ km} - 1048 \text{ km} = 48 \text{ km} = 48000 \text{ m}$
The time taken is:
$40 \text{ minutes} = 40 \times 60 \text{ s} = 2400 \text{ s}$
Therefore,the average speed is:
$V_{av} = \frac{48000 \text{ m}}{2400 \text{ s}} = 20 \text{ m/s}$
The speedometer measures the instantaneous speed of the vehicle at any given moment,not the average speed over a period of time. Therefore,the speedometer will not show this calculated average velocity.

(N/A) Given: Speed of the bus $= 60 \, km/h = 60 \times (5/18) \, m/s = 16.67 \, m/s$.
$(i)$ For a normal reaction time of $1/15 \, s$:
Distance $=$ Speed $\times$ Time $= 16.67 \times (1/15) = 1.11 \, m$.
$(ii)$ For a reaction time of $1/2 \, s$ under the influence of alcohol:
Distance $=$ Speed $\times$ Time $= 16.67 \times 0.5 = 8.335 \, m$ (approximately $8.34 \, m$).

(A) Given velocity $v = 120 \, km \, h^{-1}$.
Time $t = 30 \, s$.
First,convert time into hours: $t = 30 \, s = 30 / 3600 \, h = 1 / 120 \, h$.
Using the formula for distance: $S = v \times t$.
$S = 120 \, km \, h^{-1} \times (1 / 120) \, h = 1 \, km$.
Therefore,the train will cover a distance of $1 \, km$ in $30 \, s$.

(66 S) Length of the train $= 100 \, m = 0.1 \, km$.
Length of the bridge $= 1 \, km$.
Total distance to be covered to cross the bridge $= \text{Length of train} + \text{Length of bridge} = 0.1 \, km + 1 \, km = 1.1 \, km$.
Velocity of the train $= 60 \, km/h$.
Time taken $(t) = \frac{\text{Total distance}}{\text{Velocity}} = \frac{1.1 \, km}{60 \, km/h} = \frac{1.1}{60} \, h$.
To convert time into seconds, multiply by $3600 \, s/h$:
$t = \frac{1.1}{60} \times 3600 \, s = 1.1 \times 60 \, s = 66 \, s$.

(A) Let the distance between Ali's home and school be $x$.
During the trip to school:
Distance $= x$,Speed $= 20 \, km \, h^{-1}$.
Time taken $t_1 = \frac{x}{20} \, h$.
During the return trip:
Distance $= x$,Speed $= 30 \, km \, h^{-1}$.
Time taken $t_2 = \frac{x}{30} \, h$.
Average speed is defined as the total distance divided by the total time taken:
Average Speed $= \frac{\text{Total Distance}}{\text{Total Time}} = \frac{x + x}{t_1 + t_2} = \frac{2x}{\frac{x}{20} + \frac{x}{30}}$.
Calculating the denominator:
$\frac{x}{20} + \frac{x}{30} = \frac{3x + 2x}{60} = \frac{5x}{60} = \frac{x}{12}$.
Therefore,Average Speed $= \frac{2x}{x/12} = 2x \times \frac{12}{x} = 24 \, km \, h^{-1}$.

(D) vector quantity is a physical quantity that has both magnitude and direction.
Displacement,velocity,and acceleration are all vector quantities because they require both magnitude and direction to be fully described.
Speed is a scalar quantity because it only represents the magnitude of the rate of change of position (distance per unit time) and does not have a specific direction associated with it.
Therefore,speed is not a vector.

(B) The statement is False.
Speed is defined as the distance traveled per unit time,while velocity is defined as the displacement per unit time.
Both speed and velocity are measured in the same $SI$ unit,which is meters per second $(m/s)$.

(FALSE) The statement is False.
In one-dimensional motion with uniform velocity,the average velocity is equal to the instantaneous velocity at any point in time.
Even in non-uniform motion,the average velocity over an interval can be equal to the instantaneous velocity at a specific moment within that interval,so the statement that they are always unequal is incorrect.

(C) The momentum $(p)$ of a body is defined as the product of its mass $(m)$ and its velocity $(v)$.
Mathematically,it is expressed as $p = m \times v$.
Since the mass $(m)$ of the body is given as constant,the momentum $(p)$ is directly proportional to the velocity $(v)$ or speed of the body ($p \propto v$ when $m$ is constant).
Therefore,the momentum of a body of given mass is proportional to its speed.