23 Acceleration of a body is given by the equation
= (4 - 3v) m/s2. Speed v is in m/s and t is in
second. If initial velocity is zero then terminal
velocity of the body is
(1) 4/3 m/s
(2) 4m/s
(3) 3/4 m/s
(4) Infinite
Answers
Answered by
18
Answer:
Explanation:
Acc of the body =4-3v
Thus
dv/dt=4-3v
dv/(4-3v)=dt
Now integrating bothe sides by using substitution rule
ln(4-3v)/-3=t
ln(4-3v)=-3t
Now taking antilog on both sides
4-3v=e^(-3t)
Or 3v=4 - e^(-3t)
Now terminal velocity v=[4 - e^(-3t)]/3
robinkhardwal:
give final answer with solution
Which u write in solution i also know that already
the ans must be 4m/s
u can also calculate it by taking dv/dt=4-3v then dv=4dt-3vdt on integrating both sides u get v=4t-3s+c(integration of vdt gives v*t which is simply distance s) thus now v=0 and t=0 then c=3s thus putting the value of c in the equation gives v=4t and as the qustn suggest that t is in sec then for 1 sec v=4m/s ...
Bro answer is 4/3m/s your answer was wrong
give me correct explanation
sorry man extremely sorry the question is tooo much easy terminal velocity is that velocity in which the acceleration becomes zero and no longer pays any role in the bodys motion as in case of rainfall .
here as they asked terminal velocity then take the acc to be zero then a=4-3v now a=0 hence v=4/3m/s .......the object falls freely with constant velocity and its acc is zero that is called terminal velocity the ans that i gave at first was of the final velocity. hope now its clear
done
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