How Fast Must a Superhero Accelerate to Catch a Free-Falling Person?

[Ritchie]

Homework Statement

A hovering superhero notes a person fall past at a terminal velocity of 140 km/h. 1.9
seconds later, she begins to accelerate downwards and catches the person 1000 m
below. What was her rate of acceleration?

Homework Equations

d = v x t
d = v_ot + at²

The Attempt at a Solution

First I convert km/h to m/s for the persons velocity
140 / 3.6 = 38.88 m/s

1.9 seconds is essentially the superhero's reaction time so I find out how far the person goes in 1.9 seconds
38.88 x 1.9 = 73.88 m - but I have no idea what to do with this number

For the person to fall 1000 meters it would take
1000 / 38.88 = 25.71s

Now I isolate for a in the formula d = v_ot + at²
a = 2d/t²
a = 2000/661.22
a = 3.02 m/s²However according to the answers this answer is wrong. Can anybody help me and tell me what I'm doing wrong? I know I'm supposed to do something with that 73.88 m number but I don't know what. Can someone explain to me and tell me what it is used for?

 

[TSny]

Welcome!
The hero needs to cover 1000 m in how much time?

 

[Ritchie]

Welcome!
The hero needs to cover 1000 m in how much time?

It took 1.9 seconds for the 73.88 meters so for the rest = 1000 - 73.88 = 926.12 meters
926.12 / 38.88 = 23.81 seconds + 1.9 seconds which still gives me the same amount of time?

 

[TSny]

Suppose for the moment that the hero were to start her descent at the same instant that the person passes her. How much time does she have to catch the person at 1000 m below?

 

[mfb]

This is probably easier if you look at absolute time values.

At t=0 s the person passes the superhero.
At t=1.9 s the superhero starts accelerating
At t=? the falling person is 1000 m below the starting point.

=> the superhero accelerates for ?? seconds