The answer is d because that’s the lowest velocity and it’s a straight line showing it’s not going faster
The particle has 
and 
, and is undergoing a constant acceleration of 
.
This means its position at time 
is given by the vector function,
![\implies\vec r(t)=\left[4\,\mathrm m+\left(2\dfrac{\rm m}{\rm s}\right)t-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cimplies%5Cvec%20r%28t%29%3D%5Cleft%5B4%5C%2C%5Cmathrm%20m%2B%5Cleft%282%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29t-%5Cleft%281%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5Cright%5D%5C%2C%5Cvec%5Cimath-%5Cleft%281%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5C%2C%5Cvec%5Cjmath)
The particle crosses the x-axis when the 
component is 0 for some time 
, so we solve:
The negative square root introduces a negative solution that we throw out, leaving us with 
or about 3.24 seconds after it starts moving.
It all combines together and it freezes
<h3><u>Answer;</u></h3>
12.5 Newtons
<h3><u>Explanation;</u></h3>
Work done is defined as the product of force and distance covered or moved. It is measured in joules.
Work done = Force × distance
Therefore; making force the subject;
Force = work done/distance
= 50.0 J/ 4.00 m
=<u> 12.5 Newtons.</u>
