A particle moves a distance X=[t+5]^-1.The acceleration is directly pr...
Explanation of Proportional Acceleration
Given Information
Distance moved by the particle: X = [t 5]^-1
Solution
Step 1: Calculate Velocity
The velocity of the particle can be obtained by differentiating the distance with respect to time:
v = dx/dt = -1/[t^2*5]
Step 2: Calculate Acceleration
The acceleration of the particle can be obtained by differentiating the velocity with respect to time:
a = dv/dt = 2/[t^3*5]
Step 3: Finding the proportional relationship between acceleration and velocity or distance
The question asks for the proportional relationship between acceleration and either velocity or distance. To find this, we can substitute the expression for velocity or distance into the expression for acceleration and see which option gives a proportional relationship.
Option 1: Proportional to (velocity)^2/3
Substituting the expression for velocity into the expression for acceleration gives:
a = 2/[t^3*5] = 2/[-v^3*5*125]^(1/3)
This expression is not proportional to (velocity)^2/3, so option 1 is not correct.
Option 2: Proportional to (velocity)^3/2
Substituting the expression for velocity into the expression for acceleration gives:
a = 2/[t^3*5] = 2/[-v^3*5*125]^(1/3) = 2/[-2*5*125*5^3]^(1/3)*v^(1/3)
This expression is proportional to (velocity)^1/3, not (velocity)^3/2, so option 2 is not correct.
Option 3: Proportional to (distance)^2
Substituting the expression for distance into the expression for acceleration gives:
a = 2/[t^3*5] = 2/[X^3*5^4] = 2/[t^3*5^5*(t^2*5)^3] = 2/[t^11*5^8*X^3]
This expression is proportional to (distance)^-3, not (distance)^2, so option 3 is not correct.
Option 4: Proportional to (distance)^-2
Substituting the expression for distance into the expression for acceleration gives:
a = 2/[t^3*5] = 2/[X^3*5^4] = 2/[t^3*5^5*(t^2*5)^3] = 2/[t^11*5^8*X^3] = 2*X^2/[t^11*5^8]
This expression is proportional to (distance)^-2, so option