Derek and diane race each other on bicycles. derek starts 4 seconds before diane. he goes at a steady speed of 6 m/s. diane starts from rest and has an acceleration of 2 m/s2. a. write an expression for derek's position in terms of t and v. (1 point) b. write an expression for diane's position in terms of t and a. (1 point) c. at what time does diane catch up to derek? (2 points) d. how far has she gone when she catches up with him?

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:

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explanation:

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