Answer :
Answer:
(a) -3.173 m/s^2
(b) 3.94 s
(c) 2.47 s
Explanation:
initial velocity, u = 25 m/s
final velocity, v = 0
distance, s = 98.5 m
(a) Let a be the acceleration of the car
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
0 = 625 + 2 x a x 98.5
a = -3.173 m/s^2
(b) v = 12.5 m/s
u = 25 m/s
a = - 3.173 m/s^2
Let the time is t.
Use first equation of motion
v = u + a t
12.5 = 25 - 3.173 t
t = 3.94 s
(c) s = 98.5 / 2 = 49.25 m
u = 25 m/s
a = - 3.173 m/s^2
Let the time be t.
Let v be the velocity at this distance.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=25^{2}+2\times {-3.173}\ times 49.25tex]
v = 17.17 m/s
Use first equation of motion
v = u + at
17.17 = 25 - 3.173 x t
t = 2.47 s
