Answer: She moves 5.616 meters in that second.
Explanation:
If we define t = 0s as the moment when she starts decelerating we can write the function of acceleration as:
a(t) = -(0.63 m/s^2)
where the negative sign is because she is slowing down.
The velocity equation can be found if we integrate over time:
v(t) = -(0.63m/s^2)*t + v0
Where v0 is the constant of integration, that represents the initial velocity, in this case is:
v0 = 32.7 km/h
Now, because the acceleration is in m/s^2, we should write this velocity in m/s.
in one km we have 1000 meters, and in one hour we have 3600 seconds, then we have that:
32.7 km/h = 32.7 *(1000/3600) m/s = 9.08 m/s
Then the velocity equation becomes:
v(t) = -(0.63m/s^2)*t + 9.08 m/s
And for the position equation, we should integrate again to get:
p(t) = -(1/2)*(0.63m/s^2)*t^2 + (9.08m/s)*t + p0
Where p0 is the initial position.
For this problem, we want to find the distance that she moved between t = 5s and t = 6s, and that can be calculated as:
D = p(6s) - p(5s)
D = -(1/2)*(0.63m/s^2)*(6s)^2 + (9.08m/s)*6s + p0 +(1/2)*(0.63m/s^2)*(5s)^2 - (9.08m/s)*(5s) - p0
D = -(1/2)*(0.63m/s^2)*((6s)^2 - (5s)^2) + (9.08m/s)*(6s - 5s)
D = 5.615 m
She moves 5.616 meters in that second.
