Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = [tex]ma_a (0.35)[/tex]
[tex]1.8v^2(1.1)[/tex] - 1200(9.8)(1.25) = 1200a(0.35)
[tex]1.8v^2(1.1)[/tex] - 14700 = 420 a ------- equation (1)
[tex]F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a[/tex] --------- equation (2)
Replacing equation 2 into equation 1 ; we have :
[tex]{1.1 * 1200 \ a} - 14700 = 420 a[/tex]
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
