The graph shows the vertical velocity of a parachutist during the first 20 s of her jump.

Question 1

The graph shows the vertical velocity of a parachutist during the first 20 s of her jump.

Approximately how far does she fall before opening the parachute?

A 390 m                      B 570 m                      C 710 m                      D 770 m

Reference: Past Exam Paper – November 2016 Paper 12 Q8

Solution:

Answer: B.

Opening the parachute causes a large decrease in velocity. From the graph, the approximate time at which the parachute is opened is t = 13s (the moment BEFORE the large deceleration).

Distance travelled = area under v-t graph. So, we need to find the area from t = 0s to t = 13s. This region can be approximated as follows:

Area of triangle from t=0 to t=5: distance = ½ × 5 × 44 = 110 m

Area of trapezium from t=5 to t=10: distance = ½ × (44+60) × 5 = 260 m

Area of rectangle from t=10 to t=13: distance = 60 × 3 = 180 m

Total distance travelled = 110 + 260 + 180 = 550 m