Question

# The motion of a body is given by the equation dv(t)dt=6.0−3v(t). Where v(t) is speed in ms and t in sec. If body was at rest at t = 0

A
The terminal speed is
B
The speed varies with the time as
C
The speed is when the acceleration is half the initial value
D
The magnitude of the initial acceleration is

Solution

## The correct option is B The speed varies with the time as dvdt=6−3v⇒dv6−3v=dt Integrating both sides, ∫dv6−3v=∫dt ⇒loge(6−3v)−3=t+K1 ⇒loge(6−3v)=−3t+K2               ......(i) At t=0,v=0 ∴ loge6=K2 Substituting the value of K2 in equation (i) loge(6−3v)=−3t+loge6 ⇒loge(6−3v6)=−3t ⇒e−3t=6−3v6 ⇒v=2(1−e−3t) ∴ vterminal=2ms(whent=∞) Acceleration a=dvdt=ddt[2(1−e−3t)]=6e−3t Initial acceleration = 6ms2.Physics

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