The particle has acceleration vector
We're told that it starts off at the origin, so that its position vector at 
is
and that it has an initial velocity of 12 m/s in the positive 
direction, or equivalently its initial velocity vector is
To find the velocity vector for the particle at time 
, we integrate the acceleration vector:
![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cvec%5Cjmath)
Then we integrate this to find the position vector at time :
![\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t%2B%5Cleft%28-1.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%282.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5C%2C%5Cvec%5Cjmath)
Solve for the time when the 
coordinate is 18 m:
At this point, the coordinate is
so the answer is C.
Answer:
Towards the origin on the positive x-axis.
Explanation:
- For a negatively charged particle the direction electric field lines is always into the charge. So, when a negative charge is placed on the origin then the direction of the electric field lines will be 180° to the positive x-axis, i.e. the field lines will be travelling from infinity towards the origin.
- Contrary to this the positive charges have the direction of electric field lines emerging out from the charge.
Answer:
Velocity (v) can be calculated via v = gt, where g represents the acceleration due to gravity and t represents time in free fall. Furthermore, the distance traveled by a falling object (d) is calculated via d = 0.5gt^2.
The electric potential at a distance from a nucleus of charge 50e is 144 Volts.
<h3>What is electric potential?</h3>
The potential is the work done by the test charge to move it from infinity to a particular point.
The potential is given by the relation
V = kQ / r
The distance of a point from the nucleus is 5.0 × 10⁻¹⁰ m and the charge on the nucleus is 50 e = 50 × 1.9 × 10⁻¹⁹ C.
k = 9 × 10⁹ kg-m²/s²-C²
Here, Q is the charge, r is the distance and k is the constant.
Substituting the value into the equation and solving, we get
V = 144 Volts
Thus, the potential at a distance from a nucleus of charge 50e is 144 Volts.
Learn more about electric potential.
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