Respuesta :
Answer:
t = 6.37 s
Explanation:
If we see the motion of car with respect to truck then
relative speed of car initially and finally is ZERO
maximum relative speed of car could be
[tex]v_{max} = 50 - 35 = 15 mph[/tex]
[tex]v_{max} = 6.7 m/s[/tex]
now maximum acceleration of the car is given as
[tex]a_{max} = 5 ft/s^2 = 1.524 m/s^2[/tex]
maximum deceleration is given as
[tex]a_{max} = 20 ft/s^2 = -6.096 m/s^2[/tex]
now car has to cover the relative separation of 80 ft
[tex]d = 24.2 m[/tex]
now we have
distance covered to stop the car from maximum speed is
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 6.7^2 = 2(-6.096)d[/tex]
[tex]d_2 = 3.68 m[/tex]
time taken to cover this distance
[tex]t_2 = \frac{v}{a} = \frac{6.7}{6.096}[/tex]
[tex]t_2 = 1.1 s[/tex]
now distance cover to reach maximum speed
[tex]d_1 = \frac{v_f^2 - v_i^2}{2a}[/tex]
[tex]d_1 = \frac{6.7^2}{2(1.524)}[/tex]
[tex]d_1 = 14.7 m[/tex]
time taken to cover this distance
[tex]t_1 = \frac{v}{a} = \frac{6.7}{1.524}[/tex]
[tex]t_1 = 4.4 s[/tex]
now remaining distance is covered at constant speed
[tex]t_2 = \frac{24.2 - 3.68 - 14.7}{6.7}[/tex]
[tex]t_2 = 0.87 s[/tex]
So total time is
[tex]t = 1.1 + 4.4 + 0.87[/tex]
[tex]t = 6.37 s[/tex]
