Motion In A Plane
Class 11th Physics NCERT Exemplar Solution
Mcq I
Question 1.The angle between A =i+ j and B =i – j is
A. 45°
B. 90°
C. –45°
D. 180°
Answer:
For calculating angle between two vectors we will use the formula
Here 
therefore |A|=
similarly for |B|=
, and for dot product we will follow element-wise product of each term which will result into
Therefore 
,and this is true when 
Question 2.
Which one of the following statements is true?
A. A scalar quantity is the one that is conserved in a process.
B. A scalar quantity is the one that can never take negative values.
C. A scalar quantity is the one that does not vary from one point to another in space.
D. A scalar quantity has the same value for observers with different orientations of the axes
Answer:
We know scalar quantities like the energy of an electron can take a negative value. Also, we have an electric charge which is a scalar quantity and can be negative or positive in nature. Also in some cases where inelastic collision is happening energy is not always conserved. Electric Potential is scalar in nature but varies from one point to another in space.
But, it is always true that scalar quantity has the same value irrespective of from where we observing.
Question 3.
Figure 4.1 shows the orientation of two vectors u and v in the XY plane.
If u = aî + bĵ and
V = pî +qĵ
Which of the following is correct?
A. a and p are positive while b and q are negative.
B. a, p and b are positive while q is negative.
C. a, q and b are positive while p is negative.
D. a, b, p and q are all positive.
Answer:
When we resolve the U vector, a and b will lie on a positive x-axis and y-axis respectively, suggesting that they are positive in nature (shown in red colour in above figure)
While resolving the orange vector p lies on the positive x- axis while q lies on the negative y- axis. Therefore, a, b, p, are positive while q is negative
Question 4.
The component of a vector r along X-axis will have maximum value if
A. r is along positive Y-axis
B. r is along positive X-axis
C. r makes an angle of 45° with the X-axis
D. r is along negative Y-axis
Answer:
When vectors are resolved they have 
and
multiplied to vector’s magnitude. And we know that 
is always less than 1, this suggests that vectors’ x-component will less than the magnitude of the vector. Therefore, if a vector is lying on the x-axis, then only it will have maximum value as then 
. (Maximum value of cosθ)
Question 5.
The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be
A. 60 m
B. 71 m
C. 100 m
D. 141 m
Answer:
Given in question:
Horizontal range=50m (Assuming it to be R1)
Angle=15° (Assuming it to be θ1)
And we have to find for
Angle =45° (Assuming it to be θ2)
Also velocity is same both cases.
Relation between horizontal range R1 and θ1 is
Similarly for range R2 and θ2 it is
Dividing both equations,
Question 6.
Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are
A. Impulse, pressure and area
B. Impulse and area
C. Area and gravitational potential
D. Impulse and pressure
Answer:
We know that impulse is change in momentum per unit time and momentum is vector quantity; therefore impulse is also a vector quantity. The area is taken as vector quantity where calculation of flux is required. Area’s vector direction is normal to the its plane
Question 7.
In a two-dimensional motion, instantaneous speed v0 is a positive constant. Then which of the following are necessarily true?
A. The average velocity is not zero at any time.
B. Average acceleration must always vanish.
C. Displacements in equal time intervals are equal.
D. Equal path lengths are traversed in equal intervals.
Answer:
Here it is given speed which is scalar quantity; therefore there is no sense of direction given. Hence we can’t say displacement in equal time intervals are equal, as we don’t know whether the object is moving in a straight line or uniform circular motion
Average acceleration will be zero only when an object is travelling in a straight line, it could without changing its speed, it is moving in a circular motion, which will have non-zero acceleration
When a particle is moving in a circular path with constant speed its average velocity will be zero (displacement will zero at the starting point)
Therefore, with a surety we can only say for constant speed is that it is covering equal path lengths in equal time intervals
Question 8.
In a two-dimensional motion, instantaneous speed v0 is a positive constant. Then which of the following are necessarily true?
A. The acceleration of the particle is zero.
B. The acceleration of the particle is bounded.
C. The acceleration of the particle is necessarily in the plane of motion.
D. The particle must be undergoing a uniform circular motion
Answer:
It is given that we need to identify a necessarily true statement.
As there is no information of direction, therefore even though the particle’s speed is constant it can be accelerating. For example in a circular motion.
Acceleration depends on the velocity which in-turn depends on its magnitude and direction; therefore no condition can be applied on acceleration.
It could be that particle is moving in a straight line, therefore with a surety, we can’t say the particle is moving in a circular direction.
But, acceleration will always be in the plane of motion only.
Question 9.
Three vectors A, B and C add up to zero. Find which is false.
A. (A×B) × C is not zero unless B, C are parallel
B. (A×B).C is not zero unless B, C are parallel
C. If A, B, C define a plane, (A×B) ×C is in that plane
D. (A×B).C=|A||B||C|→ C2=A2+B2
Answer:
We have to identify false statement from above
It is given that 
Therefore taking cross product on both side
Now we know that when vectors are parallel then their cross product is zero
Taking post Cross Product on both side with C
Now this could only zero when B and C are parallel to each other as,
only when 
that’s when B and C are parallel
Therefore statement A is true.
Now taking previous equation
Taking dot product with C on both side
Now this could be zero on two conditions first is that B and C are parallel but it could be zero without C being parallel to B. As when we will take cross product of B and C, then vector perpendicular to both B and C, say vector K. And by taking dot product of K and C it will also be zero as angle between them will always be 90.Therefore B is false
Now if vector triple product of A and B and C, then vector will always lie on place which will formed by A, B and C. This could be visualized by understanding that 
will always lie in a single plane forming sides of triangle.
Now,
K will be perpendicular to plane containing A and B.
And taking cross product with C (which is also lying on same plane as that of A and B) will give a vector which is perpendicular to C but will be lying on same plane as that of A,B and C. Therefore statement C is true.
It is given in last option that 
therefore angle between vector A and vector B is 90 and we know that
, therefore 
form a triangle with angle between A and B equal to 90, therefore it is right angled triangle. Hence option D is also true.
Question 10.
It is found that |A+B|=|A|. This necessarily implies,
A. B = 0
B. A, B are antiparallel
C. A, B are perpendicular
D. A.B ≤ 0
Answer:
We have to identity statements which are always true. It is given that 
, it could be true in two conditions that is either 
, which means option (a) and (b) might true.
For forming a single condition we will multiply them, as either one of them is true it will uphold the necessary condition
We know 
(from previous equations)
Therefore their magnitude’s product will also be zero.
(This will always be true)
Therefore,
(Equality is true for B=0)
Above condition is always true
Mcq Ii
Question 1.Two particles are projected in air with speed v0 at angles θ1 and θ2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices
A. angle of projection: q1 > q2
B. time of flight: T1 > T2
C. horizontal range: R1 > R2
D. total energy: U1 > U2.
Answer:
Energy is a scalar quantity it doesn’t depend on an angle of projection; it depends only on the speed of projection, which is same for both cases.
It is given that height reached by the first particle is greater than that of second and we know that
Therefore θ1> θ2, Now for Time we know that it is directly proportional to Angle therefore, T1>T2.
For Range we can’t be sure as for complementary angle they can be equal also.
Question 2.
A particle slides down a frictionless parabolic (y = x2 ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
A. KE at P = KE at B
B. height at P = height at A
C. total energy at P = total energy at A
D. time of travel from A to B = time of travel from B to P.
Answer:
Energy is always conserved (unless it is an inelastic collision), therefore total energy at point A and P will always be equal as no energy is being lost due to friction.
But, Kinetic Energy will not always remain same, as it depends on the velocity of the particle. When Object is at Point P, its velocity will be zero (Highest point) but the velocity at B will not be equal to zero.
Height at P will be less than Height at A as the object at the highest point has zero velocity that means it is devoid of any kinetic energy, and has only potential energy. Whereas at point A, the object has partly kinetic and partly potential energy (Energy conservation law)
Time Taken from A to B will be more compared to B to P, as the distance from A to B is more compared to B to P.( Height A is more compared to Height P)
Question 3.
Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s):
A. Vav = 
[v (t1) + v (t2)]
B. Vav = 
C. r = (v (t2)) (t2 – t1)
D. aav = 
Answer:
When a particle’s acceleration is time-dependent, i.e. acceleration is non-uniform in nature then above two equations aren’t correct
Question 4.
For a particle performing uniform circular motion, choose the correct statement(s) from the following:
A. Magnitude of particle velocity (speed) remains constant.
B. Particle velocity remains directed perpendicular to radius vector.
C. Direction of acceleration keeps changing as particle moves.
D. Angular momentum is constant in magnitude but direction keeps changing.
Answer:
It is given that a particle is performing uniform motion, suggesting that particle’s speed is constant. Also from the figure, we can see that the particle’s velocity vector is always perpendicular to the radius vector. But since, particle’s direction is always changing; therefore, it is accelerating even though speed is constant. Angular momentum’s direction will always be perpendicular to the plane of the circular path.
Question 5.
For two vectors A and B, |A + B|= |A – B| is always true when
A. |A| |B| ≠ 0
B. A ⊥ B
C. |A| = |B| ≠ 0 and A and B are parallel or anti parallel
D. When either |A| or |B| is zero
Answer:
It is given that 
, it could be true when |A|=0 or |B|=0 or both are zero.
The given statement can also be true if both conditions are applied together that is
or 
Therefore we can write,
This gives the always true condition that is when A is perpendicular to B.
Vsa
Question 1.A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of 10ms-1, what is his acceleration at point R in magnitude and direction?
Answer:
It is given that cyclist maintains its constant speed =10m/s that is the magnitude of velocity is
And we acceleration’s magnitude is given by
We know the acceleration will directed from R to O since this is a centripetal acceleration it will always be towards inside.
Question 2.
A particle is projected in air at some angle to the horizontal, moves along parabola as shown in Fig. 4.4, where x and y indicate horizontal and vertical directions, respectively. Show in the diagram, direction of velocity and acceleration at points A, B and C.
Answer:
Acceleration is only due to gravity; therefore, it will always be pointing toward ground whereas velocity will be directed tangent to path taken by particle.
Question 3.
A ball is thrown from a roof top at an angle of 45° above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
(a) Greatest speed.
(b) Smallest speed.
(c) Greatest acceleration?
Answer:
Greatest speed will be experienced when ball will just reaching the bottom ground, as at time all the potential energy will be completely converted to kinetic energy hence maximum speed will be experienced
Smallest speed will be at highest point, as only single component will be present at time (horizontal), therefore at that time vector sum of velocity will be lowest.
Acceleration is only due to gravity; hence it will remain constant throughout (if we are not considering air drag as this will convert it into non-uniform motion)
Question 4.
A football is kicked into the air vertically upwards. What is its
(a) Acceleration, and (b) velocity at the highest point?
Answer:
Acceleration will always point downward, and it will be due to gravitational force.
At the highest point, ball will stop for some infinitesimally time and change its direction. Therefore at that time velocity will be zero.
Question 5.
A, B and C are three non-collinear, non-co-planar vectors. What can you say about direction of A × (B × C)?
Answer:
Let the cross product of
Therefore, K vector will be perpendicular to both B and C.
And when we will take cross product of K with A
It will be perpendicular to both K and A. Therefore, M will lie on a plane containing vector B and C
Sa
Question 1.A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.
Answer:
Let’s assume the boy throws the ball with vertical velocity uy and the velocity of the car is ux.
Therefore, the horizontal velocity of the ball is ux with respect to the boy on the footpath.
Hence, the motion with respect to the boy on the footpath is parabolic like that of a projectile
The resultant velocity of the ball is 
Angle of projection 
Question 2.
A boy throws a ball in air at 60° to the horizontal along a road with a speed of 10 m/s (36km/h). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if car has a speed of (18km/h). Give explanation to support your diagram.
Answer:
Given:
Horizontal velocity of car 
Vertical velocity of car 
Velocity of car 
Since the velocity of car is equal to the horizontal velocity of the ball, the ball only has vertical velocity with respect to the car.
Hence, the motion of the ball with respect to the car is a straight line.
Question 3.
In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.
Answer:
In the above figure, the dotted line represents the trajectory of projectile with no air resistance and continuous line represents the trajectory with air resistance.
Because of the air resistance, the horizontal velocity of the projectile is decreasing continuously and so is the energy of the particle. This makes the graph fall steeper than the rise.
Question 4.
A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?
Answer:
Given:
Height of the plane = h = 1.5km = 1500m
Horizontal velocity of the plane = u = 720km/hr = 200m/s
Let us calculate the horizontal distance between the target and the point of release of the bomb,
Now, formula for angle of sight is given by,
Question 5.
(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth’s rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? what is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2?
(b) Earth also moves in circular orbit around sun once every year with on orbital radius of 1.5 X 1011 m. What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
(Hint : acceleration 
= 
)
Answer:
(a) Given-
Radius of Earth = R = 6400km
Time period of circular motion of object around the earth = T = 1day = 24 hours
At the equator,
For an object performing circular motion acceleration can be found as,
3.38×10-2m/s2
At latitude θ,
(b) Given-
Orbital radius of Earth around Sun = 1.5×1011 m
Time period of 1 revolution = 1 year =365 days
We know acceleration during circular motion is given by,
= 5.97×10-3m/s2
Question 6.
Given below in column I are the relations between vectors a, b and c and in column II are the orientations of a, b and c in the XY plane. Match the relation in column I to correct orientations in column II.
Column I
(a) a + b = c
(b) a – c = b
(c) b – a = c
(d) a + b + c = 0
Column II
Answer:
Triangular law of addition states that if two vectors form two sides of a triangle taken in the same order then vector addition of the two vectors completes the third side of the triangle.
From this it is clear that,
In figure(i),
So, 
Option (c) matches with figure (i)
In figure(ii), 
So, 
Option (d) matches with figure (ii)
In figure(iii), 
So, 
Option (b) matches with figure (iii)
In figure(iv),
Option (a) matches with figure (iv)
Option (a) matches with figure (iv)
Option (b) matches with figure (iii)
Option (c) matches with figure (i)
Option (d) matches with figure (ii)
Question 7.
If |A| = 2 and |B| = 4, then match the relations in column I with the angle θ between A B and in column II.
Answer:
Given:
|A| = 2, |B| = 4.
(a) 
So, 
So, θ = 90°
Option (a) matches with (ii)
(b) 
So, 
So, θ = 0°
Option (b) matches with (i)
(c) 
So, 
So, θ = 60°
Option (c) matches with (iv)
(d) 
So, 
So, θ = 180°
Option (d) matches with (iii)
Question 8.
If |A|= 2 and |B| = 4, then match the relations in column I with the angle θ between A and B in column II
Answer:
Given:
|A| = 2, |B| = 4.
(a) 
So, 
So, θ = 0°
Option (a) matches with (iv)
(b) 
So, 
So, θ = 90°
Option (b) matches with (iii)
(c) 
So, 
So, θ = 30°
Option (c) matches with (i)
(d) 
So, 
So, θ = 45°
Option (d) matches with (ii)
La
Question 1.A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g =10 m/s2.
Answer:
Given:
Height of the hill = h = 500m
Speed of projection of packets = u = 125m/s
Distance of canon = 800m
Speed of canon = v = 2m/s
For time to be minimum, the packet should just pass the top of the hill and so the vertical component of velocity must be minimum.
So, 
Now let’s calculate the horizontal component of velocity.
Now, for time to reach the top of the hill,
Hence, 20 seconds is the time of flight.
Now, let’s find the horizontal distance between the canon and top of the hill.
R = uxt = 75m/s×10s =750m
Distance travelled by canon = 800m – 750m = 50m
Time taken by the canon =
Question 2.
A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R = 
.
If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig 4.5), show that it could be achieved by raising the gun to a height at least
h = ∆x
(Hint: This problem can be approached in two different ways:
(i) Refer to the diagram: target T is at horizontal distance x = R + ∆x and below point of projection y = – h.
(ii) From point P in the diagram: Projection at speed vo at an angle θ below horizontal with height h and horizontal range ∆x.)
Answer:
Speed of the shells = vo
Horizontal component of velocity of shells = vx = v0cosθ
Vertical component of velocity of shells = vy = v0sinθ
Maximum range = R = 
We know, maximum range is achieved at θ = 45°
Time of flight = t = 
Using kinematic equation,
-h = v0sinθt – gt2
Substituting the value of t,
-h = v0sinθ
– 
Here, θ = 45° , hence sinθ = cosθ = 
h = 
+ 
∴h = ∆x
Question 3.
A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig. 4.6).
(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).
(b) Time of flight.
(c) β at which range will be maximum.
(Hint: This problem can be solved in two different ways:
(i) Point P at which particle hits the plane can be seen as intersection of its trajectory (parabola) and straight line. Remember particle is projected at an angle (α + β) w.r.t. horizontal.
(ii) We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) in two different components, gx along the plane and gy perpendicular to the plane. Now the problem can be solved as two independent motions in x and y directions respectively with time as a common parameter.)
Answer:
Let’s consider the X-axis parallel to the inclined surface and Y-axis perpendicular to it. v be the speed of the projectile.
Range=R, x component of velocity = vcosβ
y component of velocity = vsinβ
x component of acceleration = gsinα
y component of acceleration = gcosα
t is the time of flight
Lets first find part b.
(b) Displacement in y direction is 0.
By 2nd kinematic equation in y-direction,
(a) By 2nd kinematic equation in x-direction,
Substituting the value of t in the above equation.
(c) In order to find the value of 
for which range is maximum,
We need to maximize 
Now,
For 
to be maximum, , 
should be maximum and the maximum value of sin(θ) is 1.
Therefore,
∴ 
Question 4.
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed vo and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.
(Hint: (i) After rebound, particle still has speed Vo to start.
(ii) Work out angle particle speed has with horizontal after it rebounds.
(iii) Rest is similar to if particle is projected up the incline.)
Answer:
Let’s consider the X-axis parallel to the inclined surface and Y-axis perpendicular to it. v be the speed of the projectile.
Range=R, x component of velocity = vsinθ
y component of velocity = vcosθ
x component of acceleration = gsinθ
y component of acceleration = gcosθ
There is no displacement in the y direction.
By 2nd kinematic equation in y-direction,
Now let’s calculate the distance along the plane the particle hit = R,
By 2nd kinematic equation in x-direction,
Substituting the value of t in the above equation.
Question 5.
A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?
(Hint: Assume north to be î direction and vertically downward to be − ĵ. Let the rain velocity vr be a î + b ĵ. The velocity of rain as observed by the girl is always vr – vgirl. Draw the vector diagram/s for the information given and find a and b. You may draw all vectors in the reference frame of ground based observer.)
Answer:
Let’s assume the velocity of rain to be 
Let the speed of the girl be vgirl
Relative velocity of Rain with respect to the girl is given by,
In the first case,
vgirl = 5m/s
= 
The rain falls vertically downwards with respect to the girl, so horizontal component of vrel = 0
∴ a = 5m/s
In the second case,
vgirl = 10m/s
= 
= (-5m/s) î + bĵ
The rain falls at an angle of 45° with respect to the girl, so the horizontal and the vertical components of vrel are equal.
∴ b = -5m/s
∴ 
Question 6.
A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s (Fig. 4.8).
(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?
(b) If he wants to start from point A on south bank and reach opposite point B on north bank,
(a) which direction should he swim?
(b) what will be his resultant speed?
(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?
Answer:
Given:
Speed of the river = vr = 3m/s
Speed of the man = vm = 4m/s
(a)
The river flows East and the man swims north
So, resultant speed of the man,
Resultant direction of the man,
(b)
If the man wants to reach the opposite point on the bank, then the horizontal component of his resultant velocity must be zero.
So, vmx = -vr = -3m/s
Now, direction in which the man should swim can be found by,
(c) Time taken by man to swim can be found by,
In case a,
In case b,
Clearly, T1 < T2
So, the man will reach the opposite bank in shorter time in case a.
Question 7.
A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle θ to the horizontal, find
(a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) what will be time of flight?
(c) what is the distance (horizontal range) from the point of projection at which the ball will land?
(d) find θ at which he should throw the ball that would maximise the horizontal range as found in (iii).
(e) how does θ for maximum range change if u >vo , u = vo , u < vo?
(f) how does θ in (v) compare with that for u = 0 (i.e.45o )?
Answer:
Given:
The initial horizontal velocity of the ball with respect to the fielder = vocosθ
The initial vertical velocity of the ball with respect to the fielder = vosinθ
The final horizontal velocity of the ball with respect to the spectator = vx = u + vocosθ
The final vertical velocity of the ball with respect to the spectator = vy = vosinθ
(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator can be found as,
tanθ = 
(b) Let t be the time of flight.
The displacement in y direction is 0.
(c) Let R be the horizontal displacement.
(d) For R to be maximum, 
should be equal to 0.
(e) When u = vo,
When, u << vo
We can assume,
u << vo, so we can say that
When, u >> vo
We can assume,
(f) When u = 0,
=45
Question 8.
Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates A = Axî + Ayĵ where î and ĵ are unit vector along x and y directions, respectively and Ax and Ay are corresponding components of (Fig. 4.9). Motion can also be studied by expressing vectors in circular polar co-ordinates as A = Ar
+ Aɵ
where = 
= cosθ î + sin θ ĵ and 
= − sin θ î +cos θ ĵ are unit vectors along direction in which ‘r’ and ‘θ ’ are increasing.
(a) Express î and ĵ in terms of and 
.
(b) Show that both ř and θ are unit vectors and are perpendicular to each other.
(c) Show that 
(ř) = ω where and 
= −ωř
(d) For a particle moving along a spiral given by r= aθ 
, where a = 1 (unit), find dimensions of ‘a’.
(e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.
Answer:
(a) We know,
(1)
(2)
Multiplying the equation (1) with sin
and (2) with cos
Adding the above two equations,
Substituting 
in equation (1),
(b) In order to show that 
and 
are perpendicular, lets show that their dot product is zero.
Therefore, and are perpendicular.
(c)
Differentiating on both sides,
(d) 
Writing dimensions on both sides,
(e) We know,
Here, a=1
Differentiating on both sides to find velocity,
Differentiating on both sides to find acceleration,
Question 9.
A man wants to reach from A to the opposite corner of the square C (Fig. 4.10). The sides of the square are 100 m. A central square of 50m × 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?
Answer:
When the man walks through sand the shortest path will be,
APQC.
Time taken to travel through sand=T1=
Here 
The shortest path outside sand will be ARC
Here,
m
m
Time taken to travel outside sand = T2=
Now, T1 < T2
Hence, the smallest value of v is 0.81m/s