A Ford Focus with a cross-sectional area of 3.00m^2 is driving on a highway at a speed of 88.0 km/hr, then accelerates to 113 km/hr to pass another car. How much more force does the motor have to supply to overcome air resistance at the higher speed? Assume the density of the air is traveling through is 1.12kg/m^3. Coefficient of drag is .32.
Ive been trying to solve this question for 6 hrs!!
To calculate the force required to overcome air resistance at the higher speed, we can use the equation:
F = 0.5 * ρ * A * Cd * v^2
Where:
F is the force
ρ is the density of the air
A is the cross-sectional area
Cd is the coefficient of drag
v is the velocity
Given:
A = 3.00 m^2
ρ = 1.12 kg/m^3
Cd = 0.32
Initial speed, v1 = 88.0 km/hr
Final speed, v2 = 113 km/hr
First, we need to convert the speeds from km/hr to m/s:
v1 = 88.0 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 24.4 m/s
v2 = 113 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 31.4 m/s
Next, we can calculate the force at the initial speed, F1:
F1 = 0.5 * ρ * A * Cd * v1^2
And the force at the final speed, F2:
F2 = 0.5 * ρ * A * Cd * v2^2
Now, we can subtract F1 from F2 to find the additional force required:
Additional force = F2 - F1
Let's calculate it step by step:
1. Calculate F1:
F1 = 0.5 * 1.12 kg/m^3 * 3.00 m^2 * 0.32 * (24.4 m/s)^2
2. Calculate F2:
F2 = 0.5 * 1.12 kg/m^3 * 3.00 m^2 * 0.32 * (31.4 m/s)^2
3. Calculate the additional force:
Additional force = F2 - F1
Now you can use these formulas and plug in the values to calculate the additional force.