A student is running at her top speed of 5.4 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.5 m from the bus, it starts to pull away, moving with a constant acceleration of 0.171 m/s2.
a) For how much time and what distance does the student have to run at 5.4m/s before she overtakes the bus?
b) When she reaches the bus how fast is the bus traveling
c) Sketch an x-t graph for both the student and the bus. Take x=0 at the initial position of the student
d) the equations you used in part a to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue thier specified motion. Explain the significance of this second solution. How fast is the bus traveling at this point
e) If the students top speed is 3.5 m/s will she catch the bus?
f) is the minumun speed the student must have to just catch up with the bus? For what time and distance must she run in that case

a) The motion of both the bus and the student can be explained by the equation [tex]x= v_{0} +\frac{1}{2}at^{2}[/tex]. Since the student is not accelerating, but rather maintaining a constant speed; the particular equation that describes the motion of the student is: [tex]x_{student} = 5.4 \frac{m}{s} * t[/tex]. Meanwhile, since the bus starts its motion at an initial velocity of zero, the equation that describes its motion is: [tex]x_{bus} =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2}[/tex]. The motion of the student relative to that of the bus can be described by the equation: [tex]x_{s} =x_{bus} +38.5m[/tex]. By replacing terms in the last equation we end up with the following quadratic equation: [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(5.4\frac{m}{s} *t)+38.5m =0[/tex]. Solving the quadratic equation will yield two solutions; t1=8.19s and t2=55.0s. By plugging in t1 onto the equation that describes the motion of the student we will find the distance runned by her, x=44.2m.

b) The velocity of the bus can be modeled by the equation [tex]v^{2} = v_{0} ^{2} +2ax[/tex]. Since the initial velocity of the bus is zero, the first term of the equation cancels. Next, we solve for v and plug in the acceleration of 0.171 m/s2 and the distance 5.74m (traveled by the bus, note: this is equal to the 44.2m travelled by the student minus the 38.5m that separated the student from the bus at the beginning of the problem).

c) The equations that make up the x-t graph are: [tex] x = 5.4 \frac{m}{s} * t[/tex] and [tex]x =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2} + 38.5[/tex]; as described in part a.

d) The first solution states that the student would have to run 44.2 m in 8.19 s in order to catch the bus. But, at that point, the student has a greater speed than that of the bus. So if both were to keep he same specified motion, the student would run past the bus until it reaches a velocity greater than that of the student. At which point, the bus will start to narrow the distance with the student until it finally catches up with the student 55 seconds after both started their respective motion.

e) No, because the quadratic equation [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(3.5\frac{m}{s} *t)+38.5m =0[/tex] has no solution. This means that the two curves that describe the distance vs time graph for both student and bus do not intersect.

f) This answer is reached by finding b of the quadratic equation. The minimum that b can be in order to find a real answer to the quadratic equation is found by solving [tex]b^{2} -4ac=0[/tex]. If we take a = 0.0855 and c = 38.5, then we find that the minimum speed that the student has to run at is 3.63 m/s. If we then solve the quadratic equation [tex](0.0855 \frac{m}{s^{2} } * t^{2} )-(3.63\frac{m}{s} *t)+38.5m =0[/tex],we will find that the time the student will run is 21.2 seconds. By pluging in that time in the equation that describes her motion: [tex]x_{student} = 3.63 \frac{m}{s} * t[/tex] we find that she has to run 77 meters in order to catch the bus.