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Zorluk: OrtaExtrapolation and Trend Prediction

Aerodynamic drag force (FdF_d) acts on vehicles as they move through the air. A group of students measured the drag force, in newtons (N\text{N}), acting on a scale model of a sports car in a wind tunnel at various wind velocities (vv), in meters per second (m/s\text{m/s}). The data from their trials are recorded in the table below:

Velocity (vv, m/s\text{m/s})Drag Force (FdF_d, N\text{N})
10101212
20204848
3030108108
4040192192

Based on the trend shown in the table, what is the expected aerodynamic drag force, in newtons (N\text{N}), acting on the scale model when the wind velocity is 50 m/s50\text{ m/s}?

Cevap: 300 N

Cevap

The expected aerodynamic drag force is 300 N.
The correct answer is 300 N because the drag force scales quadratically with velocity according to the relation Fd=0.12v2F_d = 0.12 v^2. Plugging in v=50 m/sv = 50\text{ m/s} yields 0.12×2500=300 N0.12 \times 2500 = 300\text{ N}.

Adım Adım Çözüm

1
Calculate the ratio of drag force to the square of the velocity for the given data points.
For all data points, Fd/v2=0.12F_d / v^2 = 0.12. This establishes the quadratic trend Fd=0.12v2F_d = 0.12 v^2.
Identifying the mathematical relationship between the variables is necessary to accurately extrapolate beyond the measured data range.
2
Substitute the target velocity of 50 m/s50\text{ m/s} into the identified quadratic formula.
Fd=0.12×(50)2=300 NF_d = 0.12 \times (50)^2 = 300\text{ N}.
Applying the mathematical trend allows for the calculation of the drag force at the extrapolated velocity.

Anahtar Kavram

Extrapolation of a quadratic relationship between velocity and aerodynamic drag force.
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