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melamori03 [73]
3 years ago
12

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Physics
1 answer:
Savatey [412]3 years ago
4 0
We are given the equation:

<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,    
                             x =53.24 meters

At t=2.95 sec,   
                             x =95.73 meters

Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s )  = 56.65 m/s

</span>For (b)

At t=2.20 sec,    
                             x =53.24 meters

At t=2.40 sec,   
                             x =63.36 meters

Velocity = (63.36 meters - 53.24<span> meters) / (2.40 s - 2.20 s )  = 50.6 m/s</span>
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q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

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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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