General Physics
dv/dt=4v+8 then what is rate of change of acceleration
dv/dt=4v+8 then what is rate of change of acceleration
1 Answer
Hitesh Soni
dv/dt = 4v + 8
dv/dt = 4(v + 2)
dv / (v + 2) = 4dt
applying calculus and solving differntial equation, we get
ln(v + 2) = 4t + C
since at t = 0, v = 0, thus we get
ln(0 + 2) = 0 + C
C = ln2
applying calculus and solving differntial equation, we get
ln(v + 2) = 4t + C
since at t = 0, v = 0, thus we get
ln(0 + 2) = 0 + C
C = ln2
Hence the equation is:
ln(v + 2) = 4t + ln2
ln(v + 2) – ln2 = 4t
ln[(v + 2) /2] = 4t
ln(v + 2) – ln2 = 4t
ln[(v + 2) /2] = 4t
(v + 2)/2 = e4t
v + 2 = 2e4t
v = 2e4t – 2
v + 2 = 2e4t
v = 2e4t – 2
now, acceleration, a = dv/dt
hence differentiate above equation
a = 8e4t
again differentiate
da/dt = 32e4t
hence differentiate above equation
a = 8e4t
again differentiate
da/dt = 32e4t
Since we need to know initial rate of change, thus we put t = 0
da/dt (initial) = 32e0 = 32
da/dt (initial) = 32e0 = 32
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