a.
Let's call:
the time calculated from the moment Diane starts her motion and the time calculated from the moment Derek starts his motion. Derek starts his motion 4 seconds before Diane: this means that has an "advantage" of 4 seconds, so we can write
Then:
is the uniform speed of Derek
Derek is moving in a uniform motion, so Derek's position will be given by (assuming that the starting position is zero):
We see that this is correct: in fact, when t=0 (instant when Diane starts her motion), Derek has already travelled for
b.
t is the time calculated from the moment Diane starts her motion. Diane is moving by accelerated motion, with constant acceleration and initial velocity , so its position at time t is given by the law of uniform accelerated motion:
c. t = 8.74 s
The time t at which Diane catches up with Derek is the time t at which the positions of the two persons is equal:
By solving, we have:
Which has two solutions:
t = -2.74 s --> negative, we can discarde it
t = 8.74 s --> this is our solution
d. 76.4 m
When Diane catches Derek, at t=8.74 s, she has covered the following distance:
We can verify that Derek is at the same position: