Page 90 - Class-11-Physics-Part-1_Neat

76 PHYSICS

= v 0 t + 1 t a 2 Given x (t) = 84 m, t = ?
2 2
5.0 t + 1.5 t = 84 ⇒⇒ ⇒⇒ ⇒ t = 6 s
2
1 2 At t = 6 s, y = 1.0 (6) = 36.0 m
Or, r = r + v t + at (4.34a)
0
0
2 dr
) +
ˆ
Now, the velocity v = = (5.0 + 3.0t i ˆ 2.0t j
It can be easily verified that the derivative of dt
dr At t = 6 s, v = 23.0 i +12.0 j
Eq. (4.34a), i.e. gives Eq.(4.33a) and it also
dt
2
2
satisfies the condition that at t=0, r = r . speed = v = 23 + 12 ≅ 26 m s − 1 t
o .
Equation (4.34a) can be written in component
form as 4.9 RELATIVE VELOCITY IN TWO
DIMENSIONS
1 2
x = x + v t + a t The concept of relative velocity, introduced in
ox
x
0
2 section 3.7 for motion along a straight line, can
1 be easily extended to include motion in a plane
y = y + v t + a t 2 (4.34b) or in three dimensions. Suppose that two objects
0
oy
2 y
A and B are moving with velocities v and v B
A
One immediate interpretation of Eq.(4.34b) is that (each with respect to some common frame of
the motions in x- and y-directions can be treated reference, say ground.). Then, velocity of object
independently of each other. That is, motion in A relative to that of B is :
a plane (two-dimensions) can be treated as two v = v – v (4.35a)
separate simultaneous one-dimensional AB A B
motions with constant acceleration along two and similarly, the velocity of object B relative to
perpendicular directions. This is an important that of A is :
result and is useful in analysing motion of objects v BA = v – v A
B
in two dimensions. A similar result holds for three Therefore, v AB = – v BA (4.35b)
dimensions. The choice of perpendicular and, v AB = v BA (4.35c)
directions is convenient in many physical
situations, as we shall see in section 4.10 for t Example 4.6 Rain is falling vertically with
–1
projectile motion. a speed of 35 m s . A woman rides a bicycle
–1
with a speed of 12 m s in east to west
direction. What is the direction in which
Example 4.5 A particle starts from origin she should hold her umbrella ?
at t = 0 with a velocity 5.0 î m/s and moves
t
in x-y plane under action of a force which Answer In Fig. 4.16 v represents the velocity
r
produces a constant acceleration of of rain and v , the velocity of the bicycle, the
b
$ $ $ $ $
$ $ $ $ $
2
(3.0i+2.0j) m/s . (a) What is the woman is riding. Both these velocities are with
respect to the ground. Since the woman is riding
y-coordinate of the particle at the instant a bicycle, the velocity of rain as experienced by
its x-coordinate is 84 m ? (b) What is the
speed of the particle at this time ?
Answer From Eq. (4.34a) for r0= 0, the position

of the particle is given by
r ( ) t = v 0 t + 1 t a 2
2
)(
= 5.0 t + (1/2 3.0 +i ˆ 2.0j ˆ ) t 2
ˆ
i
2 ˆ
= (5.0t + 1.5t 2 ˆ 1.0t j
) +i
Fig. 4.16
Therefore, x ( ) t = 5.0t + 1.5t 2 her is the velocity of rain relative to the velocity
y ( ) t = + 1.0t 2 of the bicycle she is riding. That is v = v – v b
r
rb
2018-19