THEORY: Relative velocity: Let us consider and
are the velocity of two particles A and B creates angleθ. The velocity of A with respect to B is
=
–
. The angle between
and –
is (180- θ) and
is the resultant of
and –
.
The magnitudes of is VAB 2 = VA 2 +VB 2 + 2VAVB cos(180-θ)
Or, VAB 2 = VA 2 +VB 2 + 2VAVB cosθ.
If creates angle φ with
then, φ =
Or, tanφ = .
CONCEPTS ON RELATIVE VELOCITY IN ONE DIMENSION:
Let us consider speed of each bus is v and the distance between the nearest two busses plying on either side is vT.
For bus going from town A to B:
Speed of bus with respect to cyclist is (v-u). As the cyclist notices that a bus goes pass him every t1 minutes in the direction of his motion, then the separation between the busses is (v-u)t1. So, vT = (v-u)t1 —- (i)
For bus going from town B to A:
Speed of bus with respect to cyclist is (v+u). As the cyclist notices that a bus goes pass him every t2 minutes in the opposite direction of his motion, then the separation between the busses is (v+u)t2. So, vT = (v+u)t2—- (ii)
From equations (i) and (ii) (v-u)t1 = (v+u)t2
Or, v = .
Using equation (i) vT = (v-u)t1
∴T = .
Muzzle speed is the speed of bullet with respect to the van. Speed of van or velocity of thief’s car are measured with respect to ground.
So, velocity of bullet with respect to van = velocity of bullet with respect to ground – velocity of van with respect to ground
W = velocity of bullet with respect to ground – u
Or, velocity of bullet with respect to ground = w + u.
Now, velocity of bullet with respect to thief’s car = velocity of bullet with respect to ground – velocity of thief’s car with respect to ground
So, velocity of bullet with respect to thief’s car = w + u – v.
At the instant when driver of car B, decides to overtake A the velocities of car A, B and C are respectively VA = u, VB = 2u and VC = -2u [ -ve due to opposite direction].
Velocity of car B with respect to car A is VBA = VB – VA = 2u – u = u.
Velocity of car C with respect to car A is VCA = VC – VA = -2u – u = -3u.
Time required by C to cross A is t = =
.
To avoid accident, car B must overtake car A in this time. Therefore, d = VBAt + at2
Or, d = u() +
a(
)2
So, a = .
CONCEPTS ON RELATIVE VELOCITY IN ONE DIMENSION:
The velocity of one train relative to the other is u – (-u) = 2u. The collision time is 2u = 2ut (s = vt) or, t = 1h. So bird moves for 1h. Now the velocity of bird with respect to train towards which it is moving will be 2u – (-u) = 3u kmh-1.
So, the time taken by bird for 1st trip, t1 = () =
h.
In this time the trains have moved towards each other 2u× =
km, so the remaining distance = 2u –
=
So, the time taken by bird for 2nd trip t2 =×
=
.
Similarly, time taken by the bird for nth trip is tn = .
Therefore, +
+ —– +
= 1
Or, [
]
Or,1- = 1
Or, = ∞
∴ n = ∞.
So, the bird makes infinite trips.
Distance travelled by bird for 1st trip is s1 = 2ut1 = 2u ×=
km.
Distance travelled by bird for2nd trip is s2 = 2ut2 = 2u× =
km.
Distance travelled by bird for nth trip is sn = 2utn = 2u× =
km.
So, total distance travelled by bird forinfinite trip is s = +
+
+ ——∞
= [1 +
+
+ ——–∞]
= [
] = 2u.
The velocity of man with respect to ground is
= ucos
– usin
.
The velocity of rain with respect to man is = – v
.
We know that =
–
so,
=
+
= – v
+ ucos
– usin
= ucos
– (usin
+v)
Therefore, actual velocity of rain is and angle with vertical is
=
(
)
The velocity of man with respect to ground at any instant of time t is = at [v = at].
The velocity of rain with respect to ground is = u.
The velocity of rain with respect to man is =
–
.
At any instant rain appears to the man at an angle , so,tan
=
.
We have to find the rate at which the angle of the axis of umbrella with vertical should be changed so that the rain falls parallel to the axis of umbrella. Therefore, we need .
=
(
)
Or, =
Or, =
=
=
=
.
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