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Svetllana [295]
3 years ago
5

Distinguish

Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

Velocity and speed both are continuously increasing.

Acceleration is constant.

Explanation:

Speed is defined as length of path covered by a body per unit time. Speed is a scalar quantity that consist of magnitude only and not direction.

Velocity is defined as the displacement per unit times. Displacement is the shortest distance between the two points. It is a vector quantity and  hence has a direction in the direction of displacement along with its own magnitude.

  • Both velocity and speed have same unit of measure which is meter per second in S.I. During <em>free fall</em> in the absence of any air resistance the velocity and speed both will be having a vertical downward direction with continuously increasing magnitude. Tough we are not concerned about the direction when discussing about speed but here both are equal since the motion is linear.

Acceleration is the rate of change in velocity of a body which is a vector quantity. For speed we are concerned about instantaneous acceleration since for a short period of time it may have a specific direction.

  • During free fall the acceleration is of a body is equal to the acceleration due to gravity and constant when the height of fall is much lesser than the radius of the earth.
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q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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