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Distance versus Displacement Acceleration
Total length of path (ACB) covered by the particle is called dv d
a= = (vx iˆ + v y ˆj + vz kˆ)
distance. Displacement vector or displacement is the minimum dt dt
distance (AB) and directed from initial position to final position.
C dvx ˆ dv y ˆ dvz ˆ
= i+ j+ k = ax iˆ + a y ˆj + az kˆ
dt dt dt
A B
Important Points About 1D Motion
Displacement is Change of Position Vector Distance ≥ | displacement | and Average speed ≥ | average
From DOAB ∆r = rB − rA velocity |
If distance > | displacement | this implies
rB = x2 iˆ + y2 ˆj + z2 kˆ
(a) atleast at one point in path, velocity is zero.
and rA = x1iˆ + y1 ˆj + z1kˆ Differentiation Differentiation
Displacement Velocity Acceleration
∆r= ( x2 − x1 )iˆ + ( y2 − y1 ) ˆj + ( z2 − z1 )kˆ Integration Integration
A
Motion with Constant Acceleration:
Dr
Equations of Motion
rA B In vector form
v= u + at and
rB
u +v 12 12
O ∆r = r2 − r1 , s = t = ut + at = vt − at
2 2 2
Displacement ∆r
Average velocity = ⇒ vav = a
Time interval ∆t v2 = u2 + 2a.s and snth = u + (2n − 1)
2
Distance travelled (Snth → displacement in nth second)
Average speed =
Time interval
In scalar form (for one dimensional motion):
For uniform motion
u+v 1 2 1 2
Average speed = | average velocity | = | instantaneous velocity| v = u + at s =
t =+
ut at =−
vt at
2 2 2
dr d ˆ ˆ a
Velocity v = = ( xi + yj + zkˆ) v2 = u2 + 2as sn = u + (2n – 1)
dt dt 2
dx ˆ dy ˆ dz ˆ
= i+ j + k = vx iˆ + v y ˆj + vz kˆ Uniform Motion
dt dt dt
If an object is moving along the straight line covers equal distance
Total change in velocity ∆v
AverageAcceleration = = a=
av in equal interval of time, it is said to be in uniform motion along
Total time taken ∆t
a straight line.
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