Respuesta :
Answer
given,
initial speed = 18 m/s
distance, d = 130 m
reaction time, t_r = 0.50 s
a) distance traveled in the reaction time
d= s x t_r
d=18 x 0.5
d = 9 m
distance of the traffic light, d = 130 - 9 = 121 m
b) deceleration
[tex]a = \dfrac{v^2-u^2}{2s}[/tex]
[tex]a = \dfrac{0^2-18^2}{2\times 121}[/tex]
a = -1.34 m/s²
c) Using equation of motion
v = u + a t
0 = 18 - 1.34 t
t = 13.43 s
Total time, T = 13.43 + 0.5 = 13.93 s
Part A. The distance from the intersection is 121 m.
Part B. The acceleration required is 1.34 m/s2
Part C. The total time to stop at the signal is 13.93 s.
How do you calculate the distance, acceleration and time?
Given that the initial speed is 18 m/s. The distance is 130 m from an intersection when the traffic light turns red. The rejection time is 0.50 s.
Part A
The distance traveled in rejection time is given below.
distance traveled in rejection time = speed [tex]\times[/tex] rejection time
[tex]d_r = s\times t_r[/tex]
[tex]d_r = 18\times 0.50[/tex]
[tex]d_r = 9\;\rm m[/tex]
Hence the distance from the traffic light is given below.
[tex]d = 130 -d_r[/tex]
[tex]d = 130 -9[/tex]
[tex]d = 121\;\rm m[/tex]
The distance from the intersection is 121 m.
Part B
The acceleration can be calculated as given below.
[tex]a = \dfrac {v^2 - u^2}{2s}[/tex]
Where u is the initial speed, v is the final speed and s is the distance from the intersection.
[tex]a = \dfrac {0^2 - (18)^2}{2\times 121}[/tex]
[tex]a = - 1.34 \;\rm m/s^2[/tex]
The acceleration required is 1.34 m/s2. Here negative sign shows that this is deacceleration.
Part C
The time to stop at the signal can be calculated as given below.
[tex]v = u +at[/tex]
[tex]0 = 18 - 1.34 \times t[/tex]
[tex]t = \dfrac {18}{1.34}[/tex]
[tex]t = 13.43 \;\rm s[/tex]
Total Time = [tex]13.43 + 0.5[/tex]
Total Time = 13.93 s
Hence the total time to stop at the signal is 13.93 s.
To know more about speed and acceleration, follow the link given below.
https://brainly.com/question/14363745.
