Answer:
a) d = 15.6m
b) v= 26.44m/s
Explanation:
Distance = 26.0m
Traveling rate = 110km/hr
=( 110*1000)/ 3600
= 30.5 m/s
Deceleration (-a) = 5.20 m/s^2
Let xi =initial position
xf = final position
Vi= initial velocity
Vf = final Velocity
a) Assume that my car starts at position xi = 0
The police car starts at circle = 26m
The initial distance between both cars = 26 - 0 = 26.0m
The position of my car after I gained my focus is given by
xf = xi + Vi(t)
= 0 + (30.5)2
= 61m
The position of the police car after I gained my focus is given as
xf = xi + Vi(t) + 1/2at^2
= 26 + (30.5)2 + 1/2(-5.2)(2^2)
= 26 + 61 - 10.4
= 87 - 10.4
= 76.6m
The separation distance after I regard my focus is 76.6 - 61
= 15.6m
D = 15.6m
b) Suppose it took another 0.50 secs to realize danger and begin to press brake, the initial position of my car (xi) = 61m
The initial position of the police car(xi) = 76.6m
The distance I traveled is given by
xf= xi + Vi(t)
= 61 + (30.5)0.5
= 61 + 15.25
= 76.25
Velocity the police car reached while decelerating when I wasn't focusing is given as
Vf= Vi + at
= 30.5 + (-5.2)2
= 30.5 - 10.4
= 20.1 m/s
This velocity is now the initial velocity of the police car. The position of the police car after 2 secs is given as
xf = xi + Vi(t) + 1/2at^2
= 76.6 + (20.1)0.5 + 1/2(-5.2)(0.5^2)
= 76.6 + 10.05 - 0.65
= 87.1 - 0.65
= 86.45m
The distance between the two cars became 86.45 - 76.25
= 10.2m
Velocity of the police car became
Vf = Vi +at
= 20.1 + (-5.2)0.5
= 20.1 - 2.6
= 17.5m/s
To know the time at which the two cars collide, set the position of the two cars equal to each other
xf(1) = xf(2)
xi(1) + Vo(t) + 1/2at^2 = xi(2) + Vi(t) + 1/2at^2
xi(1) + Vo(t) = xi(2) + Vi(t)
76.25 + (30.5)t = 86.45 + (17.5)t
30.5t - 17.5t = 86.45 - 76.25
13t = 10.2
t = 10.2/13
t = 0.78secs
Velocity of my car is given by
V= Vi +at
= 30.5 + (-5.2)(0.78)
= 30.5 - 4.056
= 26.44 m/s