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

Question 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, and angular velocity.

Solution:

Scalar: Volume, mass, speed, density, number of moles, angular frequency
Vector: Acceleration, velocity, displacement, angular velocity

Question 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, and relative velocity.

Solution:

Work and current are scalar quantities. (Although current has magnitude and direction, it does not obey the laws of vector addition; work is the scalar dot product of force and displacement vectors).

Question 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, and charge.

Solution:

$$\text{Impulse} = \text{change in momentum} = \text{force} \times \text{time}$$
Since force and momentum are vector quantities, impulse is a vector quantity.

Question 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.

Solution:

a. Yes, the addition of two scalar quantities is meaningful only if they both represent the same physical quantity (e.g., adding mass to mass, but not mass to temperature).

b. No, the addition of a vector quantity with a scalar quantity is physically meaningless because they have different directional properties.

c. Yes, a scalar can be multiplied by a vector to yield a new vector quantity. For example, mass multiplied by velocity gives momentum ($\vec{p} = m\vec{v}$).

d. Yes, a scalar can be multiplied by another scalar, regardless of whether they have the same or different dimensions.

e. Yes, the addition of two vector quantities is meaningful only if they both represent the same physical quantity (e.g., adding force to force).

f. Yes, a component of a vector can be added to the vector itself because components and vectors of the same type possess identical dimensions and directional units.

Question 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 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.

Solution:

a. True, because the magnitude of a vector represents a numerical value (scalar) with appropriate units.

b. False, because each component of a vector has a specific direction along coordinate axes along with its magnitude, making it a vector projection.

c. False, total path length is equal to the magnitude of displacement only when the particle moves along a straight line in a single fixed direction.

d. True, because total path length (distance) is always greater than or equal to the magnitude of displacement.

e. True, three non-coplanar vectors have independent directional components along three spatial dimensions and cannot form a closed loop (triangle), meaning their vector sum can never be zero.

Question 3.6

Establish the following vector inequalities geometrically or otherwise:
a. $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$
b. $|\vec{a} + \vec{b}| \ge ||\vec{a}| – |\vec{b}||$
c. $|\vec{a} – \vec{b}| \le |\vec{a}| + |\vec{b}|$
d. $|\vec{a} – \vec{b}| \ge ||\vec{a}| – |\vec{b}||$
When does the equality sign above apply?

Solution:

a. By the triangle law of vector addition, the magnitude of the resultant vector $|\vec{a} + \vec{b}|$ forms the third side of a triangle whose other two sides have lengths $|\vec{a}|$ and $|\vec{b}|$. Since the sum of any two sides of a triangle is always greater than or equal to the third side, $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$. Equality applies when vectors $\vec{a}$ and $\vec{b}$ act along the same straight line in the same direction.

b. From triangle inequality properties, the third side of a triangle is greater than or equal to the difference of the other two sides, giving $|\vec{a} + \vec{b}| \ge ||\vec{a}| – |\vec{b}||$. Equality applies when vectors act along the same straight line in opposite directions.

c. Replacing $\vec{b}$ with $-\vec{b}$ in the sum inequality gives $|\vec{a} – \vec{b}| \le |\vec{a}| + |\vec{b}|$. Equality applies when vectors act in opposite directions along a straight line.

d. Similarly, the difference inequality yields $|\vec{a} – \vec{b}| \ge ||\vec{a}| – |\vec{b}||$. Equality applies when vectors act along the same line in the same direction.

Question 3.7

Given vectors $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$, which of the following statements are correct:
a. Vectors $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ must each be a null vector,
b. The magnitude of vector $(\vec{a} + \vec{c})$ equals the magnitude of vector $(\vec{b} + \vec{d})$,
c. The magnitude of $\vec{a}$ can never be greater than the sum of the magnitudes of $\vec{b}, \vec{c},$ and $\vec{d}$,
d. Vectors $\vec{b} + \vec{c}$ 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?

Solution:

a. Incorrect, because a zero sum can be obtained when four vectors form a closed quadrilateral without individual vectors being null.

b. Correct:
$$\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0 \implies \vec{a} + \vec{c} = -(\vec{b} + \vec{d})$$
Taking modulus on both sides:
$$|\vec{a} + \vec{c}| = |-(\vec{b} + \vec{d})| = |\vec{b} + \vec{d}|$$
c. Correct:
$$\vec{a} = -(\vec{b} + \vec{c} + \vec{d}) \implies |\vec{a}| = |\vec{b} + \vec{c} + \vec{d}| \le |\vec{b}| + |\vec{c}| + |\vec{d}|$$
d. Correct: Rewriting as $\vec{a} + (\vec{b} + \vec{c}) + \vec{d} = 0$, three vectors can form a closed triangle only if the vector $(\vec{b} + \vec{c})$ lies in the same plane as $\vec{a}$ and $\vec{d}$ (or along their line if collinear).

Question 3.8

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

Solution:

$$\text{Radius of the ground } r = 200\text{ m}$$
$$\text{Magnitude of displacement} = \text{Diameter} = 2 \times r = 2 \times 200 = 400\text{ m}$$
The magnitude of displacement for each girl is equal to $400\text{ m}$. For the girl who skates along the straight path (diameter PQ), this displacement is equal to the actual length of the path skated.

Question 3.9

A cyclist starts from the centre O of a circular park of radius $1\text{ km}$, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO. If the round trip takes $10\text{ minutes}$, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Solution:

a. Since the cyclist returns to the starting point (centre O), net displacement $= 0$.

b. Average velocity $= \frac{\text{Net Displacement}}{\text{Total Time}} = \frac{0}{10\text{ min}} = 0$.

c. Total path length $= \text{OP} + \text{Arc PQ} + \text{QO} = 1\text{ km} + \left(\frac{1}{4} \times 2\pi(1)\right)\text{ km} + 1\text{ km} = 2 + \frac{\pi}{2} \approx 3.57\text{ km}$.
$$\text{Time taken} = \frac{10}{60}\text{ h} = \frac{1}{6}\text{ h}$$
$$\text{Average speed} = \frac{\text{Total Path Length}}{\text{Total Time}} = \frac{3.57\text{ km}}{1/6\text{ h}} \approx 21.42\text{ km h}^{-1}$$

Question 3.10

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

Solution:

The path forms a regular hexagon with each side equal to $500\text{ m}$.
Third turn: Displacement $= 500 + 500 = 1000\text{ m}$, Total path length $= 3 \times 500 = 1500\text{ m}$.
Sixth turn: Motorist returns to starting point. Displacement $= 0$, Total path length $= 6 \times 500 = 3000\text{ m}$.
Eighth turn: Displacement corresponds to two sides of the hexagon at an angle, magnitude $\approx 866.03\text{ m}$, Total path length $= 8 \times 500 = 4000\text{ m}$.

Question 3.11

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

Solution:

$$\text{Total distance} = 23\text{ km}, \quad \text{Displacement} = 10\text{ km}, \quad \text{Time} = \frac{28}{60}\text{ h}$$
a. $\text{Average speed} = \frac{\text{Total distance}}{\text{Time}} = \frac{23}{28/60} \approx 49.29\text{ km h}^{-1}$.

b. $\text{Average velocity magnitude} = \frac{\text{Displacement}}{\text{Time}} = \frac{10}{28/60} \approx 21.43\text{ km h}^{-1}$.

No, average speed and average velocity are not equal because the path taken is not a straight line.

Question 3.12

The ceiling of a long hall is $25\text{ m}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\text{ m s}^{-1}$ can go without hitting the ceiling of the hall?

Solution:

Maximum height $H = 25\text{ m}$, Initial speed $u = 40\text{ m s}^{-1}$.
Using maximum height formula:
$$H = \frac{u^2 \sin^2\theta}{2g} \implies 25 = \frac{(40)^2 \sin^2\theta}{2 \times 9.8} \implies \sin^2\theta = \frac{490}{1600} = 0.30625 \implies \sin\theta \approx 0.5534 \implies \theta \approx 33.6^\circ$$
Now, maximum horizontal range $R$:
$$R = \frac{u^2 \sin(2\theta)}{g} = \frac{(40)^2 \sin(67.2^\circ)}{9.8} = \frac{1600 \times 0.922}{9.8} \approx 150.53\text{ m}.$$

Question 3.13

A cricketer can throw a ball to a maximum horizontal distance of $100\text{ m}$. How much high above the ground can the cricketer throw the same ball?

Solution:

Maximum horizontal range $R_{\text{max}} = \frac{u^2}{g} = 100\text{ m}$ (at $\theta = 45^\circ$).
When thrown vertically upwards ($\theta = 90^\circ$), maximum height achieved $H$ is:
$$H = \frac{u^2}{2g} = \frac{100}{2} = 50\text{ m}.$$

Question 3.14

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

Solution:

$$\text{Radius } r = 80\text{ cm} = 0.8\text{ m}, \quad \text{Frequency } \nu = \frac{14}{25}\text{ Hz}$$
$$\text{Angular frequency } \omega = 2\pi\nu = 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25}\text{ rad s}^{-1}$$
$$\text{Centripetal acceleration } a_c = \omega^2 r = \left(\frac{88}{25}\right)^2 \times 0.8 \approx 9.91\text{ m s}^{-2}$$
The direction of centripetal acceleration is always directed along the string towards the center of the circle.

Question 3.15

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

Solution:

$$\text{Radius } r = 1000\text{ m}, \quad v = 900\text{ km h}^{-1} = 900 \times \frac{5}{18} = 250\text{ m s}^{-1}$$
$$a_c = \frac{v^2}{r} = \frac{(250)^2}{1000} = 62.5\text{ m s}^{-2}$$
$$\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38 \implies a_c = 6.38g.$$

Question 3.16

Read each statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

Solution:

(a) False, because net acceleration has a tangential component as well when speed is non-uniform.
(b) True, instantaneous velocity is always tangent to the trajectory.
(c) True, because in uniform circular motion, radial acceleration vectors pointing in opposite directions cancel out symmetrically over a complete cycle.

Question 3.17

The position of a particle is given by $\vec{r} = 3.0t\hat{i} – 2.0t^2\hat{j} + 4.0\hat{k}\text{ m}$ where $t$ is in seconds.
(a) Find the velocity ($\vec{v}$) and acceleration ($\vec{a}$) of the particle.
(b) What is the magnitude and direction of velocity of the particle at $t = 2.0\text{ s}$?

Solution:

(a) $\vec{v} = \frac{d\vec{r}}{dt} = 3.0\hat{i} – 4.0t\hat{j}\text{ m s}^{-1}$
$\vec{a} = \frac{d\vec{v}}{dt} = -4.0\hat{j}\text{ m s}^{-2}$

(b) At $t = 2.0\text{ s}$, $\vec{v} = 3.0\hat{i} – 8.0\hat{j}$:
$$\text{Magnitude } |\vec{v}| = \sqrt{(3.0)^2 + (-8.0)^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54\text{ m s}^{-1}$$
$$\text{Direction } \theta = \tan^{-1}\left(\frac{-8.0}{3.0}\right) \approx -69.45^\circ\text{ (below the } x\text{-axis)}.$$

Question 3.18

A particle starts from the origin at $t = 0\text{ s}$ with a velocity of $10.0\hat{j}\text{ m s}^{-1}$ and moves in the $x\text{-}y$ plane with a constant acceleration of $(8.0\hat{i} + 2.0\hat{j})\text{ m s}^{-2}$.
a. At what time is the $x$-coordinate of the particle $16\text{ m}$? What is the $y$-coordinate at that time?
b. What is the speed of the particle at that time?

Solution:

a. Along $x$-axis: $x = u_x t + \frac{1}{2}a_x t^2 \implies 16 = 0 + \frac{1}{2}(8.0)t^2 \implies 4t^2 = 16 \implies t = 2.0\text{ s}$.
Along $y$-axis at $t = 2.0\text{ s}$: $y = u_y t + \frac{1}{2}a_y t^2 = (10.0 \times 2.0) + \frac{1}{2}(2.0)(2.0)^2 = 20 + 4 = 24\text{ m}$.

b. Velocity components at $t = 2.0\text{ s}$:
$v_x = u_x + a_x t = 0 + (8.0 \times 2.0) = 16\text{ m s}^{-1}$
$v_y = u_y + a_y t = 10.0 + (2.0 \times 2.0) = 14\text{ m s}^{-1}$
$$\text{Speed} = \sqrt{v_x^2 + v_y^2} = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.26\text{ m s}^{-1}.$$

Question 3.19

What is the magnitude and direction of the vector $\vec{A} = 2\hat{i} + 3\hat{j}$, and what are its components along the directions of $(\hat{i} + \hat{j})$ and $(\hat{i} – \hat{j})$?

Solution:

1. Magnitude $|\vec{A}| = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.61$.
2. Direction $\theta = \tan^{-1}\left(\frac{3}{2}\right) \approx 56.31^\circ$ from $+x$-axis.
3. Components along $(\hat{i} + \hat{j})$ and $(\hat{i} – \hat{j})$: Let $\vec{A} = a(\hat{i} + \hat{j}) + b(\hat{i} – \hat{j})$.
Comparing coefficients: $a + b = 2$ and $a – b = 3 \implies 2a = 5 \implies a = \frac{5}{2}$, $b = -\frac{1}{2}$.

Question 3.20

For any arbitrary motion in space, which of the following relations are true:
a. $\vec{v}_{\text{avg}} = \frac{1}{2}(\vec{v}(t_1) + \vec{v}(t_2))$
b. $\vec{v}_{\text{avg}} = \frac{\vec{r}(t_2) – \vec{r}(t_1)}{t_2 – t_1}$
c. $\vec{v}(t) = \vec{v}(0) + \vec{a}t$
d. $\vec{r}(t) = \vec{r}(0) + \vec{v}(0)t + \frac{1}{2}\vec{a}t^2$
e. $\vec{a}_{\text{avg}} = \frac{\vec{v}(t_2) – \vec{v}(t_1)}{t_2 – t_1}$

Solution:

(a) False, holds true only for uniform acceleration.
(b) True, universal definition of average velocity vector.
(c) False, valid only for constant acceleration.
(d) False, valid only for constant acceleration.
(e) True, universal definition of average acceleration vector.

Question 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.

Solution:

a. False, energy is a scalar but not conserved in inelastic collisions.
b. False, temperature is a scalar that can be negative.
c. False, path length is a scalar with dimensions of length.
d. False, gravitational potential is a scalar that varies with position.
e. True, scalar values remain invariant under coordinate axis rotations.

Question 3.22

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

Solution:

Height $h = 3400\text{ m}$, total angle $= 30^\circ \implies$ half angle $\theta = 15^\circ$, time $t = 10\text{ s}$.
$$\text{Horizontal distance covered in } 5\text{ s} = h \tan(15^\circ) = 3400 \times 0.2679 = 910.99\text{ m}$$
$$\text{Total distance covered in } 10\text{ s} = 2 \times 910.99 = 1822\text{ m}$$
$$\text{Speed of aircraft} = \frac{1822\text{ m}}{10\text{ s}} \approx 182.2\text{ m s}^{-1}.$$

Additional Question 1

Rain is falling vertically with a speed of $30\text{ m s}^{-1}$. A woman rides a bicycle with a speed of $10\text{ m s}^{-1}$ in the north to south direction. What is the direction in which she should hold her umbrella?

Solution:

$$\text{Velocity of rain } \vec{v}_r = -30\hat{j}, \quad \text{Velocity of cyclist } \vec{v}_c = -10\hat{j}\text{ (south)}$$
$$\text{Relative velocity of rain w.r.t. woman } \vec{v}_{rc} = \vec{v}_r – \vec{v}_c = -30\hat{j} – (-10\hat{j}) = -20\hat{j} \text{ (or similar vector resolution)}.$$
Angle with vertical $\theta = \tan^{-1}\left(\frac{10}{30}\right) = \tan^{-1}\left(\frac{1}{3}\right) \approx 18^\circ$ towards the south.

Additional Question 2

A man can swim with a speed of $4.0\text{ km/h}$ in still water. How long does he take to cross a river $1.0\text{ km}$ wide if the river flows steadily at $3.0\text{ km/h}$ and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Solution:

$$\text{Time to cross width} = \frac{\text{Width}}{\text{Swimming speed}} = \frac{1\text{ km}}{4\text{ km h}^{-1}} = 0.25\text{ h} = 15\text{ minutes}$$
$$\text{Drift distance downstream} = \text{River speed} \times \text{Time} = 3\text{ km h}^{-1} \times 0.25\text{ h} = 0.75\text{ km} = 750\text{ m}.$$

Additional Question 3

In a harbor, wind is blowing at the speed of $72\text{ km/h}$ and the flag on the mast of a boat anchored in the harbor flutters along the N-E direction. If the boat starts moving at a speed of $51\text{ km/h}$ to the north, what is the direction of the flag on the mast of the boat?

Solution:

The relative velocity of wind with respect to the moving boat determines the fluttering direction. Calculations show the flag flutters almost due east ($\approx 45.11^\circ$ or $0.11^\circ$ with the east direction).

Additional Question 4

A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors $\vec{a}$ and $\vec{b}$ at different locations in space necessarily have identical physical effects?

Solution:

– Free vectors do not have a fixed location in space (position vectors do).
– Yes, vectors can vary with time (e.g., velocity or acceleration of an accelerating particle).
– No, equal vectors at different locations do not necessarily produce identical physical effects (e.g., two equal forces acting at different points on a rigid body produce different torque effects).

Additional Question 5

A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

Solution:

No. A quantity having magnitude and direction must also obey the commutative law of vector addition to be a vector (e.g., electric current has magnitude and direction but is scalar). Finite rotations do not obey commutative laws of addition, hence finite rotations are not vectors, though infinitesimal rotations are.

Additional Question 6

Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere?

Solution:

(a) No, length of a curved wire loop has no unique direction vector.
(b) Yes, a plane area can be represented by an area vector normal to the surface.
(c) No vector can be associated with the volume of a sphere, though directional area elements can be assigned to its surface.

Additional Question 7

A bullet fired at an angle of $30^\circ$ with the horizontal hits the ground $3.0\text{ km}$ away. By adjusting its angle of projection, can one hope to hit a target $5.0\text{ km}$ away?

Solution:

$$R = \frac{u^2 \sin(2\theta)}{g} \implies 3.0 = \frac{u^2 \sin(60^\circ)}{g} \implies \frac{u^2}{g} = \frac{3.0}{\sin(60^\circ)} = 2\sqrt{3} \approx 3.46\text{ km}$$
The maximum possible range ($R_{\text{max}}$ at $\theta = 45^\circ$) is $3.46\text{ km}$. Therefore, one cannot hit a target $5.0\text{ km}$ away with the same muzzle speed.

Additional Question 8

A fighter plane flying horizontally at an altitude of $1.5\text{ km}$ with speed $720\text{ km/h}$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed $600\text{ m s}^{-1}$ to hit the plane? What minimum altitude should the pilot fly to avoid being hit?

Solution:

$$\text{Plane speed } v = 720\text{ km h}^{-1} = 200\text{ m s}^{-1}, \quad \text{Muzzle speed } u = 600\text{ m s}^{-1}$$
$$\sin\theta = \frac{v}{u} = \frac{200}{600} = \frac{1}{3} \approx 0.33 \implies \theta \approx 19.5^\circ\text{ from the vertical}$$
$$\text{Minimum safe altitude} = H_{\text{max}} = \frac{u^2 \cos^2\theta}{2g} = \frac{(600)^2 \cos^2(19.5^\circ)}{2 \times 10} \approx 16000\text{ m} = 16\text{ km}.$$

Additional Question 9

A cyclist is riding with a speed of $27\text{ km/h}$. As he approaches a circular turn on the road of radius $80\text{ m}$, he applies brakes and reduces his speed at the constant rate of $0.50\text{ m/s}$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Solution:

$$v = 27\text{ km h}^{-1} = 7.5\text{ m s}^{-1}, \quad r = 80\text{ m}$$
$$\text{Centripetal acceleration } a_c = \frac{v^2}{r} = \frac{(7.5)^2}{80} \approx 0.70\text{ m s}^{-2}$$
$$\text{Tangential deceleration } a_t = 0.50\text{ m s}^{-2}$$
$$\text{Net acceleration } a = \sqrt{a_c^2 + a_t^2} = \sqrt{(0.70)^2 + (0.50)^2} = \sqrt{0.74} \approx 0.86\text{ m s}^{-2}$$
$$\text{Direction } \phi = \tan^{-1}\left(\frac{a_c}{a_t}\right) = \tan^{-1}\left(\frac{0.70}{0.50}\right) \approx 54.46^\circ\text{ with velocity direction}.$$

Additional Question 10

a. Show that for a projectile, the angle between the velocity and the $x$-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} – gt}{v_{0x}}\right)$.
b. Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4H}{R}\right)$.

Solution:

a. At any time $t$, horizontal velocity $v_x = v_{0x}$ and vertical velocity $v_y = v_{0y} – gt$. Therefore, $\tan\theta = \frac{v_y}{v_x} = \frac{v_{0y} – gt}{v_{0x}} \implies \theta(t) = \tan^{-1}\left(\frac{v_{0y} – gt}{v_{0x}}\right)$.

b. Maximum height $H = \frac{u^2 \sin^2\theta_0}{2g}$ and Range $R = \frac{u^2 \sin(2\theta_0)}{g} = \frac{2u^2 \sin\theta_0 \cos\theta_0}{g}$.
$$\frac{H}{R} = \frac{u^2 \sin^2\theta_0 / 2g}{2u^2 \sin\theta_0 \cos\theta_0 / g} = \frac{\sin\theta_0}{4\cos\theta_0} = \frac{\tan\theta_0}{4} \implies \theta_0 = \tan^{-1}\left(\frac{4H}{R}\right)$$.