PROBLEM 5Imagine that a man is running at his top speed v to catch a bus that is stopped at a traffic light. When he is still adistance d from the bus, the light changes and the bus starts to move away from the running man with aconstant acceleration a. When will the man catch the bus? Explain your answer.This is the distance travelled by the bus = dAliosof theSpeed of the personThe total distance travelled by a man to catch the bus = d+ d.consider man:Given:x, =d+d, distance travelled by the manconsider man:d = distance of the man to the busx, =v•d+d, = v•, → eq. 1before the bus start to moveconsider bus:v = velocity of the mant = time of mand, = v1+-a ·1, distance travelled by the busconsider bus:1,= time of the bus=a-t → cq. 2d, = distance travelled by the bussubstitute eq.2 to eq.1Vi = 0 velocity initial of the busa = acceleration of the busd+a•1, =v•t, , =1,note: time of the man is same of the busd+Reg:t =? time to catch the bus++d =0[ 22v±2,(v – 2ada t - 2v·1+2d = 02ause quadratic eq. to solve for t.vt(v - 2ad)ax? + bx +c = 02aa = ab = -2vc = 2dFrom here, we are only concern about-bt Vb? - 4acthe nature of the discriminant. If it is2aimaginary.v2 - 2ad → should not be less than zeroor else we will have an imaginary value.real-(-2v)± (-2v)² – 4(a)(2d)2(a)2vtV4v - 8ad2aThus, we can conclude that the man can catch the bus2v± 4(v² – 2ad)only if he will be able to attain a speed which will be2aequal to or higher that the value of sqrt(2ad).14Science is the belief in the ignorance of experts – Richard Feynman

Name: PROBLEM 5Imagine that a man is running at his top speed v to catch a
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Description: PROBLEM 5Imagine that a man is running at his top speed v to catch a bus that is stopped at a traffic light. When he is still adistance d from the bus, the light changes and the bus starts to move away from the running man with aconstant acceleratio

Suppose the man is running at his top speed (v) to catch a bus that is stopped at a traffic light.…

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istance d from the bus, the light changes and the bus starts to move away from the running man w onstant acceleration a. When will the man catch the bus? Explain your answer. This is the distance travelled by the bus = d, Amelematian of kas a Altien f te ke Spred af the person = V

istance d from the bus, the light changes and the bus starts to move away from the running man w onstant acceleration a. When will the man catch the bus? Explain your answer. This is the distance travelled by the bus = d, Amelematian of kas a Altien f te ke Spred af the person = V

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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PROBLEM 5
Imagine that a man is running at his top speed v to catch a bus that is stopped at a traffic light. When he is still a
distance d from the bus, the light changes and the bus starts to move away from the running man with a
constant acceleration a. When will the man catch the bus? Explain your answer.
This is the distance travelled by the bus = d
Aliosof the
Speed of the person
The total distance travelled by a man to catch the bus = d+ d.
consider man:
Given:
x, =d+d, distance travelled by the man
consider man:
d = distance of the man to the bus
x, =v•
d+d, = v•, → eq. 1
before the bus start to move
consider bus:
v = velocity of the man
t = time of man
d, = v1+-a ·1, distance travelled by the bus
consider bus:
1,= time of the bus
=a-t → cq. 2
d, = distance travelled by the bus
substitute eq.2 to eq.1
Vi = 0 velocity initial of the bus
a = acceleration of the bus
d+a•1, =v•t, , =1,
note: time of the man is same of the bus
d+
Reg:t =? time to catch the bus
++d =0
[ 2
2v±2,(v – 2ad
a t - 2v·1+2d = 0
2a
use quadratic eq. to solve for t.
vt(v - 2ad)
ax? + bx +c = 0
2a
a = a
b = -2v
c = 2d
From here, we are only concern about
-bt Vb? - 4ac
the nature of the discriminant. If it is
2a
imaginary.
v2 - 2ad → should not be less than zero
or else we will have an imaginary value.
real
-(-2v)± (-2v)² – 4(a)(2d)
2(a)
2vtV4v - 8ad
2a
Thus, we can conclude that the man can catch the bus
2v± 4(v² – 2ad)
only if he will be able to attain a speed which will be
2a
equal to or higher that the value of sqrt(2ad).
14
Science is the belief in the ignorance of experts – Richard Feynman

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