Answer:
Time taken, t = 3 s
Step-by-step explanation:
It is given that,
Initial velocity of the particle, u = 50 m/s
Final velocity, v = 20 m/s
Distance covered, s = 105 m
Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :
![v^2-u^2=2as](https://img.qammunity.org/2020/formulas/physics/middle-school/kzr98dbu2wfj2ipzjwf8lasb185fsfra2y.png)
![a=(v^2-u^2)/(2s)](https://img.qammunity.org/2020/formulas/physics/college/nnih5mejyth9cl990unq25l2z7yus5r04b.png)
![a=((20\ m/s)^2-(50\ m/s)^2)/(2* 105\ m)](https://img.qammunity.org/2020/formulas/physics/college/tvyl9x4tcxso42i02xktzncn7uyii0pz2c.png)
![a=-10\ m/s^2](https://img.qammunity.org/2020/formulas/physics/college/6ffzfbu2ror06jzd7bvm8b6dqatxrdo89p.png)
So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :
![t=(v-u)/(a)](https://img.qammunity.org/2020/formulas/physics/college/x99fd00llnhqqwna9ostz55ku5q1qgcrew.png)
![t=(20\ m/s-50\ m/s)/(-10\ m/s^2)](https://img.qammunity.org/2020/formulas/physics/college/o7j1qggxzuj40hxpnau6oe9bxailk2wu9i.png)
t = 3 s
So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.