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
