Question : A particle moves with a deceleration proportional to √v. Initial velocity is v₀.If the time after which it will stop is n√v₀/k then find n. [Given that 'k' is constant of proportionality.
Doubt by Suhani
Solution :
a∝-√v
a=-k√v
a=-k√v
dv/dt = -k√v
dv = -k√v dt
dv/√v = -kdt
dv/√v = -kdt
dv/(v)1/2 = -kdt
(v)-1/2 dv = -kdt
(v)-1/2 dv = -kdt
Integrating both sides within proper limits
t=2√v₀/k
t=n√v₀/k
So, on comparing, we get n=2
Similar Question :
1.) A particle moves with a deceleration k√v. Initial velocity is v₀. Find the time after which it will stop
a) [√v₀]/k
b) [√v₀]/2k
c) [2√v₀]/k
d) None of these.
2.) The acceleration experience by a body after the engine is cut off, is given by dv/dt = -kv³, where k is a constant. If v₀ is the magnitude of the velocity at cut off, find the magnitude of the velocity at time t after the cutoff.
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1.) c) [2√v₀]/k2.) v=v₀/√[1+2ktv₀²]