Stickman is driving a dragster. He does the quarter mile in 4 seconds and at the finish line is speed is 100m/s. He deploys his parachute after he crosses the finish line and wished to slow to a speed of 10m/s in 10s. What diameter parachute does he need if the drag force is equal to -Av^2 (A is the shoot area=pir^2 in m^2)
The resistive force acting on the dragster is f=-Av^2
you know the following
m=1000kg
v= speed of the dragster at any time after stickman crosses the finish line
vi=100m/s=speed of the dragster as stickman crosses the finish line
g=acceleration due to gravity
1)write newton's second law equation for the dragster any time after stickman crosses the finish line.
2) solve newton's second law equation to determine an expression for the area of the parachute attacked to the dragster at any time after stickman crossest eh finish line.
3)determine the diameter of the parachute needed to slow the dragster down from 100m/s to 10m/s in 10s.
1) Newton's second law states that the sum of the forces acting on an object is equal to its mass multiplied by its acceleration. In this case, the drag force is the only force acting on the dragster after Stickman crosses the finish line. Therefore, the equation becomes:
F = ma
where F is the drag force and a is the acceleration.
2) The drag force is given as F = -Av^2, where A is the area of the parachute attached to the dragster and v is the speed of the dragster at any time after Stickman crosses the finish line. The negative sign indicates that the force opposes the motion of the dragster.
Since the acceleration is the rate of change of velocity, we can rewrite the equation as:
F = m(dv/dt)
where dv/dt is the derivative of velocity with respect to time. Rearranging the equation, we have:
dv/dt = -(Av^2)/m
Now, we need to solve this differential equation to determine an expression for the area of the parachute attached to the dragster at any time after Stickman crosses the finish line. To do so, we can integrate both sides of the equation with respect to time:
∫dv = ∫-(Av^2)/m dt
To simplify the integration, we can separate the variables:
1/v^2 dv = -A/m dt
Integrating both sides:
∫(1/v^2) dv = -∫(A/m) dt
This gives us:
-1/v = -(A/m)t + C
where C is the constant of integration.
3) Now, let's determine the diameter of the parachute needed to slow the dragster down from 100m/s to 10m/s in 10s.
From the given information, we know that the initial speed of the dragster, vi, is 100 m/s, the final speed, vf, is 10 m/s, and the time taken, t, is 10 s.
Using these values in the equation derived in step 2, we can find the constant of integration, C:
-1/vi = -(A/m) * 0 + C
C = 1/vi
Substituting the values of vi, vf, and t back into the equation, we can solve for the area of the parachute:
-1/vf = -(A/m) * t + 1/vi
-1/10 = -(A/1000) * 10 + 1/100
Simplifying the equation gives:
-1/10 = -10A/1000 + 1/100
Multiplying through by 1000:
-100 = -10A + 10
Simplifying further:
-90 = -10A
Dividing by -10:
9 = A
The area of the parachute (A) is given as pi * r^2, where r is the radius of the parachute. Therefore, A = pi * r^2.
Substituting the value of A, we have:
9 = pi * r^2
Solving for the radius (r), we get:
r^2 = 9/pi
Taking the square root:
r = √(9/pi)
Finally, to determine the diameter (D) of the parachute, we multiply the radius (r) by 2:
D = 2 * √(9/pi)