KinematicsMedium
Question
A particle moves a distance x in the time t according to equation x = (t + 5)-1 . The acceleration of particle is proportional to
Options
A.(Velocity)3/2
B.(Distance)2
C.(Distance)-2
D.(Velocity)2/3
Solution
Distnace = x = (t + 5)-1 .....(i)
Velocity, V =
(t + 5)-1 = -(t + 5)-2 .....(ii)
Acceleration , a = [-(t + 5)-2] = 2 (t + 5)-3 .....(iii)
From equaton (ii), we get
= v3/2 = - (t + 5)-3 .....(vi)
Substituting this in equation (iii) and we get
Acceleration,a = - v3/2
or, a ∝ (velocty)3/2
From equation (i), we get
= x3 = (t + 5)-3
Substituting this in equation (iii) and we get
Acceleration, a = 2x3
or, a ∝ (distance)3
Hence option (a) is correct.
Velocity, V =
(t + 5)-1 = -(t + 5)-2 .....(ii)Acceleration , a =
[-(t + 5)-2] = 2 (t + 5)-3 .....(iii)From equaton (ii), we get
= v3/2 = - (t + 5)-3 .....(vi)
Substituting this in equation (iii) and we get
Acceleration,a = - v3/2
or, a ∝ (velocty)3/2
From equation (i), we get
= x3 = (t + 5)-3
Substituting this in equation (iii) and we get
Acceleration, a = 2x3
or, a ∝ (distance)3
Hence option (a) is correct.
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