can you please provide detailed solutions of question 19.

As this is a one-dimensional motion, we can use the kinematic equations. Using sign convention, we have u= 9m/sa=-2m/s^2Now the distance covered by the particle in the 5th second would be equal to the total displacement after 5s - total displacement after 4s.Now, by using the kinematic equation s=ut+1/2at^2 We obtain that the displacement at the end of 4s is 20 and at the end of 5s is also 20. Thus, the distance covered in the 5th s is 0m.Alternative method:The instantaneous velocity at the end of 4th s or the immediate beginning of 5th s would be the same and can be calculated with the help of the equation v=u+at with which we obtain v=1m/s at that instant. Now again using the equation s=ut+1/2at^2 for the time period of 1st, we get the displacement as 0m yet again. Alternative method 2 : s=(u+v) /2 *t. That is displacement is average velocity *timeThe initial velocity as calculated above is 1m/s and final velocity can be calculated similarly and equals (-1)m/s. As average velocity for the time period of 1 is 0m/s, the displacement would be 0m.*Note: This time period is for the particle to halt and change its direction. 1s might seem like a huge amount for the particle to stay at rest momentarily and change direction but as the retardation is very small(2m/s^2), 1s is not such a big amount of time. Hope this helped!!!