Question

a skydiver steps out of an airplane. the velocity-time graph shows how her velocity changes with time. (note: down is defined as the + direction of velocity.) at what points during the skydive does she experience the greatest, a zero, and the smallest (but non - zero) air resistance? tap on the fields below to toggle through the answer options. air resistance is zero at -- air resistance is greatest at -- air resistance is least (but not zero) at --

Explanation:

Step1: Recall air resistance and velocity - time graph

Air resistance (\(F_{air}\)) is related to velocity (\(v\)) by the formula \(F_{air}\propto v^{n}\) (usually \(n = 2\) for turbulent flow, but for the purpose of this analysis, we know that air resistance increases with increasing velocity. Also, the net force \(F_{net}=mg - F_{air}\), and from Newton's second law \(F_{net}=ma\), where \(a\) is the acceleration (the slope of the velocity - time graph).

At point A: The skydiver just steps out of the plane. Her velocity \(v = 0\). Since air resistance is a force that opposes motion and depends on velocity (for non - zero velocity, air resistance acts; at \(v = 0\), there is no air resistance because there is no relative motion between the skydiver and the air to cause resistance). So air resistance is zero at A.

Step2: Analyze the slope (acceleration) and air resistance

The acceleration \(a=\frac{\Delta v}{\Delta t}\), which is the slope of the velocity - time graph. The net force \(F_{net}=mg - F_{air}=ma\), so \(F_{air}=mg - ma\).

• For the greatest air resistance: We need the largest \(F_{air}\). Since \(F_{air}=mg - ma\), a smaller acceleration (closer to zero) means a larger \(F_{air}\). At point D, the velocity - time graph has a slope of zero (constant velocity, terminal velocity), so \(a = 0\). Then \(F_{air}=mg\), which is the largest possible value of air resistance (because when \(a = 0\), the air resistance balances the weight).
• For the smallest non - zero air resistance: We need the smallest \(F_{air}\) (but \(F_{air}

eq0\)). From \(F_{air}=mg - ma\), a larger acceleration (larger slope) means a smaller \(F_{air}\). At point B, the slope of the velocity - time graph is larger than at point C (and non - zero, since at C the slope is smaller than at B and larger than at D). So at point B, the acceleration \(a\) is relatively large, so \(F_{air}=mg - ma\) is relatively small (but non - zero, because \(v
eq0\) at B, so there is some air resistance).

Answer:

• Air resistance is zero at: A
• Air resistance is greatest at: D
• Air resistance is least (but not zero) at: B