Answer:
her total time taken for the race is 11.58 s
Explanation:
mass of the sprinter, m = 59 kg
acceleration of the sprinter, a = 3.9 m/s²
initial distance covered by the sprinter, d₁ = 12 m
Time taken to cover the first 12 m;
d₁ = ut + ¹/₂at²
where;
u is her initial velocity = 0
d₁ = ¹/₂at²
2d₁ = at²
[tex]t_1 = \sqrt{\frac{2d_1}{a} } \\\\t_1 = \sqrt{\frac{2\times 12}{3.9} }\\\\t_1 = 2.48 \ s[/tex]
The velocity of the sprinter at the end of the 12 m;
v = u + at
v = 0 + at
v = at
v = 3.9 x 2.48
v = 9.67 m/s
The last distance covered by the sprinter at the constant velocity of 9.67 m/s = (100 - 12)m = 88 m
The time taken to cover 88 m with an initial velocity of 9.67 m/s is calculated as;
d₂ = vt + ¹/₂at²
Since the velocity is constant, the acceleration, a = 0
d₂ = vt₂
t₂ = d₂/v
t₂ = 88 / 9.67
t₂ = 9.1 s
The total time taken for the race = t₁ + t₂
= 2.48 s + 9.1 s
= 11.58 s
