A problem of Irodov
If the expression for co-ordinates of position of particle is given, then the problem be best tackled by component method.
Question : A point moves in the plane xy such that x = a sin b t and y = a ( 1- cos b t ) ,
where a and b are +ve constants. Find :
(a) the distance S traversed by the point during the time T ;
(b) the angle between the point's velocity and acceleration vector.
solution :we shall solve the problem by component method.
the the distance S traversed by the point during the time T is given by :
S =
[ ]
[ ] | V | dt ( plz note that only speed is enough to calculate distance )
To calculate | V | we have to find
V ( bold letters mean vector )
To calculate
V , we proceed remembering
V = i V
x +
j V
y Where,
i and
j are unit vectors along X and Y axis.
Now how to calculate V
x and V
y ?
It is given, x = a sin b t => V
x = dx / dt = a b cos b t
also, y = a ( 1- cos b t ) => V
y = dy / dt = a b sin b t
so,
V = i V
x +
j V
y =
i ( a b cos b t ) +
j ( a b sin b t )
so, | V | =
[ ]
[ ( a b cos b t )
2 + ( a b sin b t )
2 ] = a b
instantaneous speed V = a b
so, S =
[ 0]
[T ] | V | dt =
[0 ]
[ T] a b dt = a b T
S = a b T
(b )
In the same component method, the acceleration of the point is calculated. The x component of acceleration is :
w
x = dV
x / dt = d { a b cos b t } / dt = -a b
2 sin b t
The y component of acceleration is :
w
y = dV
y / dt = d { a b sin b t } / dt = + a b
2 cos b t
so that acceleration
vector is
w =
i w
x +
j w
y or,
w =
i ( -a b
2 sin b t ) +
j ( a b
2 cos b t )
Now , we have velocity vector
V =
i ( a b cos b t ) +
j ( a b sin b t )
acceleration
vector
w =
i ( -a b
2 sin b t ) +
j ( a b
2 cos b t )
Let C be the angle these 2 vectors. Then
w.V = [
i ( -a b
2 sin b t ) +
j ( a b
2 cos b t ) ] . [
i ( a b cos b t )
+
j ( a b sin b t ) ]
wV cos C = - a
2 b
3 sin b t cos b t + a
2 b
3 sin b t cos b t = 0
cos C = 0
= > C = 90 degree.