Explanation:
Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR
Answer:
1. ![r(t) = (\alpha t - \frac{\beta t^3}{3})\^i + (\frac{\gamma t^2}{2})\^j](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%28%5Calpha%20t%20-%20%5Cfrac%7B%5Cbeta%20t%5E3%7D%7B3%7D%29%5C%5Ei%20%2B%20%28%5Cfrac%7B%5Cgamma%20t%5E2%7D%7B2%7D%29%5C%5Ej)
2. ![a(t) = \frac{dv(t)}{dt} = -2\beta t \^i + \gamma \^j](https://tex.z-dn.net/?f=a%28t%29%20%3D%20%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%20%3D%20-2%5Cbeta%20t%20%5C%5Ei%20%2B%20%5Cgamma%20%5C%5Ej)
3. ![r(t=2.12) = (6~{\rm m})\^j](https://tex.z-dn.net/?f=r%28t%3D2.12%29%20%3D%20%286~%7B%5Crm%20m%7D%29%5C%5Ej)
Explanation:
The velocity vector is
![v(t) = (\alpha - \beta t^2)\^i + (\gamma t)\^j](https://tex.z-dn.net/?f=v%28t%29%20%3D%20%28%5Calpha%20-%20%5Cbeta%20t%5E2%29%5C%5Ei%20%2B%20%28%5Cgamma%20t%29%5C%5Ej)
1. The position vector can be found by integrating the velocity vector.
![r(t) = \int v(t) dt = (\alpha t - \frac{\beta t^3}{3} + C_1)\^i + (\frac{\gamma t^2}{2} + C_2)\^j](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%5Cint%20v%28t%29%20dt%20%3D%20%28%5Calpha%20t%20-%20%5Cfrac%7B%5Cbeta%20t%5E3%7D%7B3%7D%20%2B%20C_1%29%5C%5Ei%20%2B%20%28%5Cfrac%7B%5Cgamma%20t%5E2%7D%7B2%7D%20%2B%20C_2%29%5C%5Ej)
At t = 0, the bird is at the origin, so the integration constants can be determined.
![r(0) = C_1 \^i + C_2 \^y = 0](https://tex.z-dn.net/?f=r%280%29%20%3D%20C_1%20%5C%5Ei%20%2B%20C_2%20%5C%5Ey%20%3D%200)
Therefore, C1 and C2 are equal to zero.
![r(t) = (\alpha t - \frac{\beta t^3}{3})\^i + (\frac{\gamma t^2}{2})\^j](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%28%5Calpha%20t%20-%20%5Cfrac%7B%5Cbeta%20t%5E3%7D%7B3%7D%29%5C%5Ei%20%2B%20%28%5Cfrac%7B%5Cgamma%20t%5E2%7D%7B2%7D%29%5C%5Ej)
2. The acceleration vector is the derivative of the velocity vector with respect to time.
![a(t) = \frac{dv(t)}{dt} = -2\beta t \^i + \gamma \^j](https://tex.z-dn.net/?f=a%28t%29%20%3D%20%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%20%3D%20-2%5Cbeta%20t%20%5C%5Ei%20%2B%20%5Cgamma%20%5C%5Ej)
3. We will first find the time when the x-component of the position vector is equal to zero.
![\alpha t - \frac{\beta t^3}{3} = 0\\\alpha = \frac{\beta t^2}{3}\\t = \sqrt{\frac{3\alpha}{\beta}} = 2.12~s](https://tex.z-dn.net/?f=%5Calpha%20t%20-%20%5Cfrac%7B%5Cbeta%20t%5E3%7D%7B3%7D%20%3D%200%5C%5C%5Calpha%20%3D%20%5Cfrac%7B%5Cbeta%20t%5E2%7D%7B3%7D%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B3%5Calpha%7D%7B%5Cbeta%7D%7D%20%3D%202.12~s)
We will plug in this value into the y-component of the position vector.
![y(t) = \frac{\gamma t^2}{2}\\y(2.12) = \frac{4(2.12)^2}{2} = 9~m](https://tex.z-dn.net/?f=y%28t%29%20%3D%20%5Cfrac%7B%5Cgamma%20t%5E2%7D%7B2%7D%5C%5Cy%282.12%29%20%3D%20%5Cfrac%7B4%282.12%29%5E2%7D%7B2%7D%20%3D%209~m)
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