Answer:
Time taken, t = 3 s
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 :
[tex]v^2-u^2=2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}[/tex]
[tex]a=-10\ m/s^2[/tex]
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 :
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}[/tex]
t = 3 s
So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.
