Respuesta :
Answer:
Explanation:
Given
For first part acceleration is [tex]a_1=2.3\ m/s^2[/tex]
time [tex]t_1=16.2\ s[/tex]
distance traveled [tex]s_1[/tex] is given by
[tex]s=ut+\frac{1}{2}at^2[/tex]
where s=displacement
u=initial velocity
a=acceleration
t=time
[tex]s_1=0+0.5\times 2.3\times (16.2)^2[/tex]
[tex]s_1=301.806\ m[/tex]
For second part speed acquired [tex]v_1[/tex] is held constant for [tex]t_2=18.4\ s[/tex]
[tex]v_1=u+a_1t_1[/tex]
[tex]v_1=2.3\times 16.2[/tex]
[tex]v_1=37.26\ m/s[/tex]
Distance traveled during this time
[tex]s_2=v_1\times t_2[/tex]
[tex]s_2=37.26\times 18.4[/tex]
[tex]s_2=685.584\ m[/tex]
For the last part it decelerate with [tex]a_3=-2.5\ m/s^2[/tex] until the car stops so
distance moved in last part is given by
[tex]v^2-u^2=2 as[/tex]
here [tex]v=0[/tex]
[tex]u=v_1[/tex]
[tex]s=s_3[/tex]
[tex]0-(37.26)^2=2\times (-2.5)\times s_3[/tex]
[tex]s_3=277.66\ m[/tex]
Total distance [tex]S=s_1+s_2+s_3[/tex]
[tex]S=1265.05\ m[/tex]
