Question

plzzz ans......plzz use equation of motion in giving ans.......Thnk u

Answer 1

distance travelled before break applied =
= 25 x5
= 125 m

retardation = 0- 25 / 10
retardation = -2.5

distance after applied break
= 25 x 10
= 250 m
_______________________#Lutz_____

Answer 2

using 2nd equation of motion
s = ut + 1/2 at²
s = 25(5) + 1/2(0)(25)
as the body is moving with uniform velocity the acceleration is zero
s = 125 + 0 = 125 m
ii) u = 25 m/s
v = 0m/s
t= 10 s
using first equation of motion
V=u+at
0 = 25+ 10a
-25/10 = a = -2.5 m/s²
iii) using 3rd equation of motion
v²= u²+2as
0²= (25)² + 2(-2.5)s
0= 625 - 5s
5s = 625
s = 625/5 = 125 m
Hope this helps