1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Bezzdna [24]
3 years ago
13

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 d

istance 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

Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

a) t=8.19s; x=44.2m

b) v=1.401 m/s

c) see attachment

d) The second solution is a later time at which the bus catches the student. v=9.40 m/s

e) No, she won't.

f) v=3.63m/s;t=21.2; x=77m

Explanation:

a) The motion of both the bus and the student can be explained by the equation x= v_{0} +\frac{1}{2}at^{2}. Since the student is not accelerating, but rather maintaining a constant speed; the particular equation that describes the motion of the student is: x_{student} = 5.4 \frac{m}{s} * t. Meanwhile, since the bus starts its motion at an initial velocity of zero, the equation that describes its motion is: x_{bus} =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2}. The motion of the student relative to that of the bus can be described by the equation: x_{s} =x_{bus} +38.5m. By replacing terms in the last equation we end up with the following quadratic equation: (0.0855 \frac{m}{s^{2} } * t^{2} )-(5.4\frac{m}{s} *t)+38.5m =0. 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 v^{2} = v_{0} ^{2} +2ax. 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: x = 5.4 \frac{m}{s} * t and  x =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2} + 38.5; 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 (0.0855 \frac{m}{s^{2} } * t^{2} )-(3.5\frac{m}{s} *t)+38.5m =0 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 b^{2} -4ac=0. 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 (0.0855 \frac{m}{s^{2} } * t^{2} )-(3.63\frac{m}{s} *t)+38.5m =0,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: x_{student} = 3.63 \frac{m}{s} * t we find that she has to run 77 meters in order to catch the bus.

You might be interested in
When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?
Oduvanchick [21]
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
a = r \times ( \frac{1}{n1}  -  \frac{1}{n2} )
We insert the values

a = 2.18 \times  {10}^{ - 18}  \times ( \frac{1}{ {1}^{2} } -  \frac{1}{ {6}^{2} }  )
= 2.12 \times {10}^{ - 18}
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
wavelength =  \frac{h \times c}{a}
Finally we insert the values
\frac{6.626 \times  {10}^{ - 34}  \times 3 \times  {10}^{8} }{2.12 \times  {10}^{ - 18} }  = 9.376 \times  {10}^{ - 8}
Which is the same as 93.8 nm
3 0
3 years ago
The shanghai maglev train is capable of reaching speeds of up to 350km h-1 what is the speed in m/h
LenaWriter [7]

Answer:

The shanghai maglev train is capable of reaching speeds of up to 217.48 in mph

217.48 mph

Explanation:

8 0
3 years ago
Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the
BlackZzzverrR [31]

relation between potential difference and electric field is given as

E . d = \Delta V

so here we know that

d = 3 cm

\Delta V = 30 V

E \times 0.03 = 30

E = 1000 N/C

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0
3 years ago
An important thing to consider when responding to a driver in front of you that stops suddenly is:a. the mental state of the oth
Gre4nikov [31]
The correct answer is option A. i.e. An important thing to consider when responding to a driver in front of you that stops suddenly is: the mental state of the other driver.

Our talk or discussion can disrturb the balance of the driver or he can get distracted. So, we must try not to speak much while the driver is driving because by doing this we are putting the life of ourselves in danger. Any distrcaction of driver can cause accident.
5 0
3 years ago
A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two
Andrew [12]
A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers. <span>The given values are
F1 = 200 N</span> F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N
F2 = 400 - 200
F2 = 200 N



4 0
3 years ago
Other questions:
  • Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of
    12·1 answer
  • The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
    15·1 answer
  • Which of the following states of matter has a define shape
    15·1 answer
  • How are the helium atoms in this model different from real helium atoms?
    14·2 answers
  • Stand on a bathroom scale on a level floor, and the reading on the scale shows the gravitational force on you, mg. If the floor
    6·1 answer
  • Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45
    7·1 answer
  • When a 4kg mass is hung vertically on a light spring, the spring stretches 2.5cm. A.) How far will the spring stretch if an addi
    12·1 answer
  • Which of the following would NOT be considered vigorous aerobic exercise?
    5·2 answers
  • Use the equation for motion to answer the question.
    10·1 answer
  • Which kind of inclined plane pushes up more? Steeper or flatter?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!