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Sagot :
(a)In the next 15.0s, she traveled 101.250 m.
(b) 4.50 m/s is the runner's final speed.
Given:- initial velocity = 9m/s , deacceleration = 0.3 m/s2
distana(d), she moves in time (t)=15 s,
and we must determine this using the second equation of motion.
S = ut + 1/2 at2,
where s = the distance
u = initial velocity
t = time
a=-d = 0.300m/s
t=15.0
By putting all values in the formula above,
we have
d = (9.00)x(15.0)-1/2 (0.300) (15.0)²
d= 101.250 m.
In the next 15.0s, she traveled 101.250 m.
final speed (v) =?
from 1 Equation of Motion : v = u+at.
To solve, v=?
u=9.00 m/s
a= -d= 0.300 m/s²
t=15.0
v=u+at
v= 9 - 0.300 X15.0
= 4.50 m/s
4.50 m/s is the runner's final speed.
Learn more about Equation of Motion here:
https://brainly.com/question/15080570
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