NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Plane

Q. 3.1 State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity

Answer:

• Volume is a scalar quantity since it has only magnitude without any direction.
• Mass is a scalar quantity because it is specified only by magnitude.
• Speed is specified only by its magnitude, not by its direction, so it is a scalar quantity.
• Acceleration is a vector quantity as it has both magnitude and direction associated.
• Density is a scalar quantity as it is specified only by its magnitude.
• The number of moles is a scalar quantity as it is specified only by its magnitude.
• Velocity is a vector quantity as it has both magnitude and direction.
• Angular frequency is a scalar quantity as it is specified only by its magnitude.
• Displacement is a vector quantity since it has both magnitude and direction.
• Angular velocity is a vector quantity as it has both magnitude and direction.

Q. 3.2: Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity

Answer:
The two scalar quantities are work and current, as these two do not follow the laws of vector addition.

Q. 3.3: Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge

Answer:
Among all, impulse is the only vector quantity as it is the product of two vector quantities (force and time). Also, it has an associated direction.

Q. 3.4: State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector

Answer:

• (a) Adding two scalars is meaningful if both have the same unit or represent the same physical quantity.
• (b) Adding a scalar to a vector of the same dimensions is meaningless because a vector has direction, while a scalar does not.
• (c) Multiplying a vector by a scalar is meaningful, as it changes only the magnitude of the vector while keeping its direction unchanged.
• (d) Multiplying two scalars is valid and meaningful without any condition. The resulting scalar may represent a different physical quantity (units get multiplied).
• (e) Adding two vectors is meaningful only if they represent the same physical quantity, allowing their magnitudes and directions to be combined appropriately.
• (f) Adding a component of a vector to the same vector is meaningful as it is a case of valid vector addition, leading to a new vector with increased magnitude in the same direction.

Q. 3.5: Read each statement below carefully and state with reasons, if it is true or false:
(a) The magnitude of a vector is always a scalar,
(b) Each component of a vector is always a scalar,
(c) The total path length is always equal to the magnitude of the displacement vector of a particle.
(d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater than or equal to the magnitude of the average velocity of the particle over the same interval of time,
(e) Three vectors not lying in a plane can never add up to give a null vector.

Answer:

• (a) True: The magnitude of a vector has no direction, so it is always a scalar.
• (b) False: Each component of a vector is itself a vector, as it retains directional properties.
• (c) False: Total path length equals the magnitude of displacement only in straight-line motion. In general, path length is ≥ displacement.
• (d) True: Average speed is total path length/time, which is ≥ average velocity (displacement/time).
• (e) True: Three vectors not lying in the same plane cannot cancel each other out, hence cannot give a null vector.

Q. 3.6 (a) Establish the following vector inequalities geometrically or otherwise:

Inequality: $|\vec{a} + \vec{b}|$ ≤ $|\vec{a}| + |\vec{b}|$

Answer:

Consider the image below:

In triangle OCB:

OB < OC + BC ⟹ $|\vec{a} + \vec{b}|$ < $\vec{|a|} + \vec{|b|}$ ..........(i)

If $\vec{a}$ and $\vec{b}$ are in a straight line:

$|\vec{a} + \vec{b}|$ = $|\vec{a}| + |\vec{b}|$ ..........(ii)

Conclusion: From (i) and (ii), $|\vec{a} + \vec{b}|$ ≤ $|\vec{a}| + |\vec{b}|$

---

Q. 3.6 (b) Establish the following vector inequalities geometrically or otherwise:

Inequality: $|\vec{a} + \vec{b}|$ ≥ $|\vec{a}| + |\vec{b}|$

Answer:

Refer to the same image above.

In triangle OCB:

OB + BC > OC ⟹ OB > |OC − BC|

Therefore, $|\vec{a} + \vec{b}|$ > $|\vec{a}| + |\vec{b}|$ ..........(i)

If vectors a and b are opposite in a straight line:

$|\vec{a} + \vec{b}|$ = $||\vec{a}| − |\vec{b}||$ ..........(ii)

Conclusion: From (i) and (ii), $|\vec{a} + \vec{b}|$ ≥ $||\vec{a}| − |\vec{b}||$

---

Q. 3.6 (c) Establish the following vector inequalities geometrically or otherwise:

Inequality: $|\vec{a} - \vec{b}|$ ≤ $|\vec{a}| + |\vec{b}|$

Answer:

Refer to the image below:

In triangle OAB:

OB < OA + AB ⟹ $|\vec{a} − \vec{b}|$ < $|\vec{a}| + |\vec{b}|$ ..........(i)

If vectors a and b are along the same line:

$|\vec{a} − \vec{b}|$ = $|\vec{a}| + |\vec{b}|$ ..........(ii)

Conclusion: From (i) and (ii), $|\vec{a}-\vec{b}|=|\vec{a}|+|\vec{b}|$

Q. 3.7 Given $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0, which of the following statements are correct:

(a) $\vec{a}, \vec{b}, \vec{c}, and \ \vec{d}$ must each be a null vector,

(b) The magnitude of $(\vec{a} +\vec{c})$ equals the magnitude of $(\vec{ b} +\vec{d})$,

(c) The magnitude of a can never be greater than the sum of the magnitudes of $\vec{b},\vec{c}, and\ \vec{d}$,

(d) $\vec{b} + \vec{c}$ must lie in the plane of $\vec{a} \ and \ \vec{d}$ if $\vec{a} \ and \ \vec{d}$ are not collinear, and in the line of $\vec{a} \ and \ \vec{d}$, if they are collinear?

Answer:

(a) Incorrect: - The sum of three vectors in a plane can be zero. So it is not a necessary condition that all of $\vec{a}$, $\vec{b}$, $\vec{c}$, $\vec{d}$ should be null vectors.

(b) Correct : We are given that $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$

So, $\vec{a} + \vec{b} = - (\vec{c} + \vec{d})$

Thus magnitude of $\vec{a} + \vec{c}$ is equal to the $\vec{c}+\vec{d}.$

(c) Correct :- We have $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$

$ \vec{b} + \vec{c} + \vec{d} = -\vec{a}$

So clearly, the magnitude of a cannot be greater than the sum of the other three vectors.

(d) Correct: - Sum of three vectors is zero if they are coplanar.

Thus, $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$

or $\vec{a} + (\vec{b} + \vec{c}) + \vec{d} = 0$

Hence $(\vec{b}+\vec{c})$ must be coplaner with $\vec{a} \ and \ \vec{d}$

Q. 3.8 Three girls skating on a circular ice ground of radius $200 m$ start from a point $P$ on the edge of the ground and reach a point $Q$ diametrically opposite to $P$ following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?

Answer: 400 m.

Explanation:

The displacement vector is defined as the shortest distance between two points which a particle has covered.

In this case, the shortest distance between these points is the diameter of the circular ice ground.

Thus, Displacement = 400 m.

Girl B had travelled along the diameter, so the path travelled by her is equal to the displacement.

Q. 3.9 (a) A cyclist starts from the centre $O$ of a circular park of radius $1\; km$, reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along $QO$ as shown in Fig. If the round trip takes $10 min,$ what is the net displacement?

Answer:

The net displacement, in this case, will be zero because the initial and final positions are the same.

Net displacement = Final position - Initial position

Q. 3.9 (b) A cyclist starts from the centre $O$ of a circular park of radius $1\; km,$ reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along $QO$ as shown in Fig. If the round trip takes $10 \; min,$ what is the average velocity?

Answer:

(b) Average velocity is defined as the net displacement per unit time. Since we have the net displacement to be zero, the average velocity will also be zero.

$Avg.\ Velocity\ =\ \frac{Net\ displacement}{Time\ taken}$

Q. 3.9 (c) A cyclist starts from the centre $O$ of a circular park of radius $1\; km,$ reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along $QO$ as shown in Fig. If the round trip takes $10 \; min,$ what is th the average speed of the cyclist?

Answer: \ 21.42\ Km/h$

Explanation:

(c) For finding the average speed, we need to calculate the total path travelled.

Total path = OP + arc PQ + OQ

$= 1+ \frac{1}{4}(2\pi \times1)+ 1$

$= 3.57\ Km$

$Time\ taken\ in\ hour\ = \frac{1}{6}$

So the average speed is :

$= \frac{3.57}{\frac{1}{6}}\ =\ 21.42\ Km/h$

Q. 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ after every $500 \; m.$ Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Answer: 4Km

Explanation:

The track is shown in the figure given below:-

Let us assume that the trip starts at point A.

The third turn will be taken at D.

So displacement will be = Distance AD = 500 + 500 = 1000 m

Total path covered = AB + BC + CD = 500 + 500 + 500 = 1500 m

The sixth turn is at A.

So the displacement will be zero

and total path covered will be = 6 (500) = 3000 m

The eighth turn will be at C.

So the displacement = AC

$AC =\ \sqrt{AB^2\ +\ BC ^2\ +\ 2(AB)(BC) \cos 60^{\circ}}$

or $AC=\ \sqrt{(500)^2\ +\ (500) ^2\ +\ 2(500)(500) \cos 60^{\circ}}$

$AC=\ 866.03\ m$

And the total distance covered = 3000 + 1000 = 4000 m=4Km

Q. 3.11 (a) A passenger arriving in a new town wishes to go from the station to a hotel located $10 \; km$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23 \; km$ long and reaches the hotel in $28 \; min.$ What is the average speed of the taxi?

Answer: \ 49.29\ Km/h$

Explanation:

(a) Avg. speed of taxi is given by:-

$=\ \frac{Total\ path\ travelled }{Total\ time\ taken}$

$=\ \frac{23\ Km}{\frac{28}{60}\ h}$

$=\ 49.29\ Km/h$

Q. 3.11 (b) A passenger arriving in a new town wishes to go from the station to a hotel located $10 \; km$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23\; km$ long and reaches the hotel in $28 \; min.$ What is the magnitude of the average velocity? Are the two equal?

Answer: \ 21.43\ Km/h$

Explanation:

Total displacement = 10 Km

Total time taken in hours :

$=\ \frac{28}{60}\ hr$

Avg. velocity :

$=\ \frac{10}{\frac{28}{60}}\ hr$

$=\ 21.43\ Km/h$

It can be clearly seen that the average. speed and avg. Velocity is not the same.

Q. 3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40m/s can go without hitting the ceiling of the hall?

Answer: \ 150.53\ m$

Explanation:

It is known that the maximum height reached by a particle in projectile motion is given by :

$h\ =\ \frac{u^2sin^2\theta }{2g}$

Putting the given values in the above equation :

$25\ =\ \frac{40^2sin^2\theta }{2\times9.8}$

So, we get

$\sin \theta \ =\ 0.5534$ and $\theta \ =\ 33.60^{\circ}$

Now the horizontal range can be found from :

$R\ =\ \frac{u^2 \sin 2\theta }{g}$

or $R=\ \frac{40^2 \sin (2\times 33.6 )}{9.8}$

or $R=\ 150.53\ m$

Q. 3.13 A cricketer can throw a ball to a maximum horizontal distance of $100\; m$. How high above the ground can the cricketer throw the same ball?

Answer: \ 50\ m$

Explanation:

We are given the range of projectile motion.

$R\ =\ \frac{u^2\ \sin 2\theta }{g}$

Substituting values :

$100\ =\ \frac{u^2\ \sin 90^{\circ} }{g}$

So, $\frac{u^2 }{g}\ =\ 100$

Now, since deceleration is also acting on the ball in the downward direction :

$v^2\ -\ u^2\ =\ -2gh$

Since the final velocity is 0, the maximum height is given by :

$H\ =\ \frac{u^2}{2g}$

or $H\ =\ 50\ m$

Q. 3.14 A stone tied to the end of a string $80\; cm$ long is whirled in a horizontal circle with a constant speed. If the stone makes $14$ revolutions in $25 \; s,$ what is the magnitude and direction of acceleration of the stone?

Answer: Fr\ 9.91\ m/s^2$

Explanation:

equency is given by :

$Frequency\ =\ \frac{No.\ of\ revolutions}{Total\ time\ taken}=\ \frac{14}{25}\ Hz$

And, the angular frequency is given by :

$\omega \ =\ 2 \pi f$

Thus, $\omega=\ 2 \times\ \frac{22}{7}\times\frac{14}{25}$

$\omega=\ \frac{88}{25}\ rad/s$

Hence, the acceleration is given by :

$a\ =\ \omega ^2 r$

or $a=\ \left ( \frac{88}{25} \right ) ^2 \times 0.8$

or $a\ =\ 9.91\ m/s^2$

Q. 3.15 An aircraft executes a horizontal loop of radius $1.00\; km$ with a steady speed of $900\; km/h.$ Compare its centripetal acceleration with the acceleration due to gravity.

Answer: \ 6.38$

Explanation:

Convert all the physical quantities in SI units.

$Speed\ =\ 900\times \frac{5}{18}\ 250\ m/s$

So the acceleration is given by :

$a\ =\ \frac{v^2}{r}$

$a=\ \frac{(250)^2}{1000}$

$a=\ 62.5\ m/s^2$

The ratio of centripetal acceleration to gravity gives :

$\frac{a}{g}=\ \frac{62.5}{9.8}\ =\ 6.38$

Q. 3.16 (a) Read each statement below carefully and state, with reasons, if it is true or false :

The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

Answer:

False:- Since the net acceleration is not directed only along the radius of the circle. It also has a tangential component.

Q. 3.16 (b) Read each statement below carefully and state, with reasons, if it is true or false :

The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

Answer:

True: - Because the particle moves on the circumference of the circle, at any point, its direction should be tangential in order to move in a circular orbit.

Q. 3.16 (c) Read each statement below carefully and state, with reasons, if it is true or false :

The acceleration vector of a particle in uniform circular motion, averaged over one cycle, is a null vector

Answer:

True: - In a uniform circular motion, acceleration is radially outward all along the circular path. So in 1 complete revolution, all the vectors are cancelled, and the null vector is obtained.

Q. 3.17 (a) The position of a particle is given by $\vec{r}=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$ where $t$ is in seconds, and the coefficients have the proper units for $r$ to be in metres. Find the $\vec{v}$ and $\vec{a}$ of the particle?

Answer: \ -4\; \hat{j}$

Explanation:

(a) We are given the position vector $\vec{r}=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$

The velocity vector is given by:-

$\vec{v} \ =\ \frac{dr}{dt}$

$\vec{v}\ =\ \frac{d\left ( 3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k} \right )}{dt}$

or $\vec{v}\ =\ 3\; \hat{i}-4t\; \hat{j}$

Now for acceleration :

$\vec{a} \ =\ \frac{dv}{dt}$

$\vec{a}=\ d(\frac{3\; \hat{i}-4t\; \hat{j})}{dt}$

$\vec{a}=\ -4\; \hat{j}$

Q. 3.17 (b) The position of a particle is given by $\vec{r}=3.0t\; \hat{i}-2.0t^{2}\; \hat{j}+4.0\; \hat{k}\; m$ where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres. What is the magnitude and direction of velocity of the particle at $t=2.0\; s ?$

Answer: \ -69.45^{\circ}$

Explanation:

Put the value of time t = 2 in the velocity vector as given below :

$\vec{v}\ =\ 3\; \hat{i}-4t\; \hat{j}$

or $\vec{v}\ =\ 3\; \hat{i}-4(2)\; \hat{j}$

or $\vec{v}=\ 3\; \hat{i}-8\; \hat{j}$

Thus, the magnitude of velocity is :

$|\vec{v}|=\ \sqrt{3^2\ +\ (-8)^2}\ =\ 8.54\ m/s$

Direction :

$\theta \ =\ \tan^{-1} \frac{8}{3}\ =\ -69.45^{\circ}$

Q. 3.18 (a) A particle starts from the origin at $t=0\; s$ with a velocity of $10.0\; \hat{j}\; m/s$ and moves in the x-y plane with a constant acceleration of $\left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}$. At what time is the x- coordinate of the particle $16 \; m?$ What is the y-coordinate of the particle at that time?

Answer: 24 m.

Explanation:

We are given the velocity of the particle as $10.0\; \hat{j}\; m/s$.

And the acceleration is given as :

$\left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}$

So, the velocity due to acceleration will be :

$\vec{a}\ =\ \vec{\frac{dv}{dt}}$

So, $\vec{dv}\ =\ \left ( 8.0\ \widehat{i}\ +\ 2.0\ \widehat{j} \right )dt$

By integrating both sides,

or $\vec{v}\ =\ 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u$

Here, u is the initial velocity (at t = 0 sec).

Now,

$\vec{v}\ =\vec{\frac{dr}{dt}}$

or $\vec{dr}\ =\ \left ( 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u \right )dt$

Integrating both sides, we get

$\vec{r}\ =\ 8.0\times \frac{1}{2}t^2\ \widehat{i}\ +\ 2.0\times\frac{1}{2} t^2\ \widehat{j}\ +\ (10t\ ) \widehat{j}$

or $\vec{r}\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}$

or $x\widehat{i}\ +\ y\widehat{j}\ =\ 4t^2\ \widehat{i}\ +\ (t^2\ +\ 10t\ ) \widehat{j}$

Comparing coefficients, we get :

$x\ =\ 4t^2$ and $y\ =\ 10t\ +\ t^2$

In the question, we are given x = 16.

So t = 2 sec

and y = 10 (2) + 22 = 24 m.

Q. 3.18 (b) A particle starts from the origin at $t=0\; s$ with a velocity of $10.0\; \hat{j}\; m/s$ and moves in the x-y plane with a constant acceleration of $\left ( 8.0\; \hat{i}+2.0\; \hat{j} \right )m\; s^{-2}.$ What is the speed of the particle at the time?

Answer: \ 21.26\ m/s$

Explanation:

The velocity of the particle is given by :

$\vec{v}\ =\ 8.0t\ \widehat{i}\ +\ 2.0t\ \widehat{j}\ +\ u$

Put t = 2 sec,

So velocity becomes :

$\vec{v}\ =\ 8.0(2)\ \widehat{i}\ +\ 2.0(2)\ \widehat{j}\ +\ 10\widehat{j}$

or $\vec{v}\ =\ 16\ \widehat{i}\ +\ 14\ \widehat{j}$

Now, the magnitude of velocity gives :

$\left ||\vec{ v}| \right |\ =\ \sqrt{16^2\ +\ 14^2}$

$|\vec{v}|=\ \sqrt{256+196}$

$|\vec{v}| =\ 21.26\ m/s$

Q 3. 19 $\hat i$ and $\hat j$ are unit vectors along the x- and y-axes, respectively. What is the magnitude and direction of the vectors $\hat i + \hat j$, and $\hat i - \hat j$? What are the components of a vector $\vec{A}=2\hat i +3 \hat j$ along the directions of $\hat i + \hat j$ and $\hat i - \hat j$? [You may use the graphical method]

Answer: \ \frac{-1}{\sqrt{2}}$

Explanation:

Let A be a vector such that:- $\overrightarrow{A}\ =\ \widehat{i}\ +\ \widehat{j}$

Then the magnitude of vector A is given by : $\left | A \right |\ =\ \sqrt{1^2\ +\ 1^2}\ =\ \sqrt{2}$

Now, let us assume that the angle made between vector A and the x-axis is $\Theta$.

Then we have:-

$\theta \ =\ \tan^{-1}\left ( \frac{1}{1} \right )\ =\ 45^{\circ}$

Similarly, let B be a vector such that:- $\overrightarrow{B}\ =\ \widehat{i}\ -\ \widehat{j}$

The magnitude of vector B is : $\left | B\right |\ =\ \sqrt{1^2\ +\ (-1)^2}\ =\ \sqrt{2}$

Let $\alpha$ be the angle between vector B and the x-axis :

$\alpha = \tan^{-1}\left ( \frac{-1}{1} \right ) =-45^{\circ}$

Now consider $\overrightarrow{C}\ =\ 2\widehat{i}\ +\ 3\widehat{j}$ :-
Then the required components of a vector C along the directions of $(\hat{i}+\hat{j})$ are:- $=\ \frac{2+3}{\sqrt{2}}\ =\ \frac{5}{\sqrt{2}}$
and the required components of a vector C along the directions of $(\hat{i}-\hat{j})$ is:- $\frac{2-3}{\sqrt{2}}\ =\ \frac{-1}{\sqrt{2}}$

Q. 3.20 For any arbitrary motion in space, which of the following relations are true :

(a) $v_{average}=\left ( 1/2 \right )\left [ v\left ( t_{1} \right )+v\left ( t_{2} \right ) \right ]$

(b) $v_{average}=\left [ r\left ( t_{2} \right ) -r\left ( t_{1} \right )\right ]/\left ( t_{2}-t_{1} \right )$

(c) $v(t)=v\left ( 0 \right )+a\; t$

(d) $v(t)=r\left ( 0 \right )+v\left ( 0 \right )t+\left ( 1/2 \right )a\; t^{2}$

(e) $a_{average}=\left [ v(t_{2})-v(t_{1}) \right ]/\left ( t_{2}-t_{1} \right )$

Answer:

(a) False:- Since it is an arbitrary motion, the following relation cannot hold for all the arbitrary relations.

(b) True:- This is true as this relation relates displacement with time correctly.

(c) False: - The given equation is valid only in the case of uniform acceleration motion.

(d) False:- The given equation is valid only in the case of uniform acceleration motion. But this is an arbitrary motion, so acceleration can be non-uniform.

(e) True:- This is the universal relation between acceleration and velocity-time, as the definition of acceleration is given by this.

Q. 3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false :

A scalar quantity is one that

(a) is conserved in a process

(b) can never take negative values

(c) must be dimensionless

(d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes

Answer:

(a) False: For e.g. energy is a scalar quantity but is not conserved in inelastic collisions.

(b) False:- For example, temperature can take negative values in degrees Celsius.

(c) False:- Since speed is a scalar quantity, it has dimensions.

(d) False:- Gravitational potential varies in space from point to point.

(e) True:- Since it doesn't have a direction.

Q. 3.22 An aircraft is flying at a height of $3400\; m$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0\; s$ apart is $30^{\circ},$ what is the speed of the aircraft?

Answer: \ 182.24\ m/s$

Explanation:

The given situation is shown in the figure:-

To find the speed of the aircraft, we just need to find the distance AC, as we are given t = 10 sec.

Consider $\Delta$ ABD,

$\tan 15^{\circ}\ =\ \frac{AB}{BD}$

$AB\ =\ \ BD\ \times \tan 15^{\circ}$

or $AC\ =\ 2AB\ =\ \ 2BD\ \times \tan 15^{\circ}$

or $=\ \ 2\times 3400\times \tan 15^{\circ}$

or $=\ 1822.4\ m$

Thus, the speed of aircraft :

$=\ \frac{1822.4}{10}\ =\ 182.24\ m/s$