Respuesta :
Answer:
18.47 s
Explanation:
Given:
Speed of the opponent, u₁ = 2 m/s
Time after which player starts = 2 seconds
Thus,
distance covered (s₁) by the opponent before the player starts = u₁ × 2
= 2 × 2 = 4 m
let 't' seconds be the time taken by the player to catch the opponent
now,
the total distance covered by the player will be
= 4 m + distance traveled by the opponent in the time t
= 4 + ( 2 × t ) m
also,
From Newton's equation of motion the distance covered by the player
[tex]s_2=ut+\frac{1}{2}at^2[/tex]
where,
s is the distance
u is the initial speed = 0 m/s as starting from rest
a is the acceleration = 0.24 m/s²
t is the time
on substituting the respective values, we get
[tex]s_2=0\times t+\frac{1}{2}\times0.24\times t^2[/tex]
also,
s₁ = s₂
thus,
[tex]4 + 2\times t=0\times t+\frac{1}{2}\times0.24\times t^2[/tex]
or
4 + 2t = 0.12t²
or
0.12t² - 2t - 4 = 0
on solving the quadratic equation, we get
t = 18.47 s
[negative value of the roots is ignored as time cannot be negative]
