Answer:
The time taken is 5.55 seconds
Explanation:
Density, p = 1kg/m³
Mass of vehicle, m = 700kg
Initial speed, u = 120m/s
Area of drag chute, A = 7.5 m²
Drag coefficient, Cd = 1.4
Final velocity, v = 20m/s
time taken to decelerate to 20m/s, t = ?
EF = ma
Since the force taken into consideration is the drag force due to the drag chute,
EF = -Fd = -(pACdv²)/2 = ma
a = dv/dt
So the equation can be written as,
-(pACdv²)/2 = m dv/dt
Taking integral of the left hand side with respect to dt and the right with respect to dv with boundaries of 0 to t and u to v,
(pACd)t/2m = (1/v) - (1/u)
t = (2m/pACd) x ( (1/v) - (1/u) ) = f(V)
At V = v
t = (2 x 700)/(1)(7.5)(1.4) x (1/20) - (1/120)
t = 5.55 seconds
