Test: Relative Velocity in Two Dimensions - JEE MCQ

We can use the Pythagorean theorem to solve this problem since the boat's velocity, river's velocity, and resultant velocity form a right triangle.

Let the boat's velocity be represented by A (10 km/hr), the river's velocity be represented by B, and the resultant velocity be represented by C (14 km/hr).

According to the Pythagorean theorem, A^2 + B^2 = C^2.

Now, we can plug in the given values and solve for B:

(10)^2 + B^2 = (14)^2
100 + B^2 = 196

Subtract 100 from both sides:

B^2 = 96

Now, take the square root of both sides:

B = sqrt(96) ≈ 9.79 km/hr

Hour hand i.e- 60x60=3600sec
Time period=2π/angular speed.
T=2π/ ω
ω=2π/60x60=2π/3600
to complete one rotation, time taken=12 hours.
For 12 hours=12x60x60 sec
=43200 sec
Angular speed of hour hand=2π/T
(i.e., a complete round) =2π/43200
= π/21600 rad/s

When the man rides, he should put an umbrella in the opposite direction to which the drops are falling with respect to him.
Hence wrt man, the speed of rain drop is
at an angle of tan^-1 ⅓
Hence he would put umbrella to the direction of tan^-1 ⅓

Radius r = 0.5 m
Speed, v = 20 m/s
We know,
v = r ω
where, 
ω = angular velocity
20 = 0.5 x ω
ω = 20/0.5 = 40 rad/sec

Correct Answer :- d

Explanation : Let the initial velocity of the body be u m/s; and the constant acceleration be a m/s². Then displacement (distance covered) of the body after t second is given by

s = u t + ½ a t²       —(1)

The body covers 50 m in 5s and 80 m in 10s.

50 = u×5 + ½ a 5²; => 50 = 5u + 25/2a

10u + 25a = 100     —(2)

80 = u×10 + ½ a 10²; => 80 = 10u + 50a; this gives,

u + 5a = 8  —(3)

Solving (2) and (3) for u and a we get

u = 12 m/s and a = -4/5 m/s²  — (4)

To obtain the distance covered in the next 15s we need to find the velocity at the beginning of this interval using,

v = u + a t => v = 12 m/s + (-4/5) m/s² × 15s = 0m/s

s = u t + ½ a t²

Substituting various values we get

s = 12×15 +½ ×(-4/5)×15² = 180 - 90

= 90 m

Velocity of boat wrt river = velocity of boat wrt ground - velocity of river wrt ground
I.e. v = (6i + 8j) - (-6i - 4j)
= 12i + 12j

The second hand goes through 2π radians in 1 min, or 2π radian/60 seconds, so ω = π/30 rad.s^-1 = 0.03 rad.s^-1

Assuming the direction of velocity of train to be positive, wrt train the speed of monkey will be
-100 + 5 = -95 kmph
I.. 95 kmph in direction opposite to the train.

Radius r = 0.5 m

Speed, v = 20 m/s

We know,

v = r ω

where, 

ω = angular velocity

20 = 0.5 x ω

ω = 20/0.5 = 40 rad/sec