Question

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?

Answer 1

a. $x(t) = vt' = v(t+4) = 6(t+4) [m]$

Let's call:

$t$ the time calculated from the moment Diane starts her motion and $t'$ 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

$t'=t+4$

Then:

$v=6 m/s$ 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):

$x(t) = vt' = v(t+4) = 6(t+4) [m]$

We see that this is correct: in fact, when t=0 (instant when Diane starts her motion), Derek has already travelled for

$x(t=0)=v(0+4)=6 (4)=24 m$

b. $x'(t)= \frac{1}{2}at^2 =\frac{1}{2}(2)t^2=t^2 [m]$

t is the time calculated from the moment Diane starts her motion. Diane is moving by accelerated motion, with constant acceleration $a=2 m/s^2$ and initial velocity $v_0=0$, so its position at time t is given by the law of uniform accelerated motion:

$x'(t)= \frac{1}{2}at^2 =\frac{1}{2}(2)t^2=t^2$

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:

$x(t)=x'(t)$

By solving, we have:

$6 (t+4) = t^2\\t^2 -6t -24 =0$

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:

$x'(t=8.74s)=t^2 = (8.74 s)^2=76.4 m$

We can verify that Derek is at the same position:

$x(t=8.74 s)=6(t+4)=6(8.74 +4)=76.4 m$