Answer:
d = 1.954 Km
Explanation:
given,
total distance, D = 2.5 Km
in stretch A to B =
speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t
in stretch B to C
time = 3.4 s
In stretch C to D
speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t
we know,
distance = speed x time
distance of BC
using equation of motion
v = u + a t
27.22 = 13.34 - a x 3.4
a = 4.08 m/s²
uniform deceleration is equal to 4.08 m/s²
distance traveled in BC


s = 68.94 m

3000 = 27.5 t + 68.94 + 13.33 t
40.83 t = 2931.06
t = 71.79 s
distance travel in AB
distance = s x t
d = 27.22 x 71.79
d = 1954 m
d = 1.954 Km
distance between A and B is equal to 1.954 Km.
Genus’s mastermind hope that helps
1) find speed (8.8 m/s)
2) find acceleration (38.7 m/s^2)
answer is about 38.7 m/s^2
As we know that reaction time will be

so the distance moved by car in reaction time



now the distance remain after that from intersection point is given by

So our distance from the intersection will be 100 m when we apply brakes
now this distance should be covered till the car will stop
so here we will have


now from kinematics equation we will have



so the acceleration required by brakes is -2 m/s/s
Now total time taken to stop the car after applying brakes will be given as



total time to stop the car is given as

Answer: i'm not fully sure but i think it's A.
Explanation: why do i think the answer is a? well the parallel circuit definition is a closed circuit in which the current divides into two or more paths before recombining to complete the circuit. which means that if you unscrew/remove one bulb it would still work, but in a series circuit it will not work, because if you remove one it turns all the bulbs off.