Andrea, a 59.0 kg sprinter, starts a race with an acceleration of 3.900 m/s2. If she accelerates at that rate for 12.00 m and then maintains that velocity for the remainder of a 100.00 m dash, what will her time (in s) be for the race

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

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

6 cos 34, 6 sin 34 trig