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If the force acting on a body is inversely proportional to its velocity, then the kinetic energy acquired by the body in time t is proportional to a.t^0 b.t^1 Explanation?
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Explanation:

The force acting on a body is given by the equation F = k/v, where F is the force, k is a constant, and v is the velocity of the body.

To find the kinetic energy acquired by the body in time t, we need to integrate the force with respect to velocity from an initial velocity v1 to a final velocity v2.

The work done on the body is equal to the change in kinetic energy, so we can write:

W = ∫(F dv) = ∫(k/v dv)

To integrate this expression, we can use u-substitution. Let u = ln(v), then du = (1/v) dv.

The integral becomes:

W = ∫(k e^u du) = k ∫(e^u du) = k e^u + C

Now, let's substitute back for u:

W = k e^ln(v) + C = k v + C

Since the work done is equal to the change in kinetic energy, we can write:

ΔKE = k v2 + C - (k v1 + C) = k(v2 - v1)

This equation shows that the change in kinetic energy is directly proportional to the change in velocity.

Now, let's consider the time t. The change in velocity can be written as:

Δv = v2 - v1 = at

where a is the acceleration.

Since a = dv/dt, we can write:

dv = a dt

Substituting this expression in the equation for the change in kinetic energy, we get:

ΔKE = k(at) = kat

This shows that the change in kinetic energy is directly proportional to the acceleration multiplied by time.

Therefore, the kinetic energy acquired by the body in time t is proportional to a.t^1.
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