# Encounter and Overtake

We know that in Uniform Motion the velocity of the mobile is constant and different from zero, while the acceleration is zero.

The time function of spaces in uniform motion is:

S = S0 + vt

Exercises

1) Two cars, A and B, move in uniform motion and in the same direction. Their speeds are respectively equal to 15 m/s and 10 m/s. At time t = 0, the cars are in the positions shown below.

Determine:
a) the time when A reaches B;
Solution:
Before starting to solve the problem we have to find the time functions of the spaces of each car.
Thus,

SA = S0A + vA.t and SB = S0B + vB.t
SA = 0 + 15.t and SB = 100 + 10.t Having

found the time functions, we have that:
to determine the instant that A meets B, it is enough to equate the functions SA = SB.

0 + 15.t = 100 + 10.t
15t – 10t = 100
5t = 100
Therefore,
t = 20 s.

b) how far from the initial position of A does the encounter occur.
To find the distance from the point A where the encounter occurs, just substitute the time value, found in the previous item, in the hourly function of the space of the car A.
Thus:

SA = 0 + 15 . 20
SA = 300 m
Therefore, the rendezvous position is 300 m from the initial position A.

2) Two cars move in uniform motion towards each other. Their scalar speeds are 12 m/s and 8 m/s, respectively. At time t = 0 the cars occupy the positions indicated in the figure:

Let’s orient the trajectory from A to B and adopt the origin of the spaces in the initial position A.
The resolution of this problem is similar to the previous one.
a) Write the time functions of the space of A and B.
The time functions of A and B are:

SA = 0 + 12.t and SB = 200 – 8t.

Here the speed of B is -8 because the car is moving in the opposite direction to the direction of the trajectory.

b) Determine the moment of the encounter:

Equating the time functions of the spaces we have:

SA = SB
12.t = 200 – 8.t
12.t + 8t = 200
20.t = 200
t = 10 s
Therefore, the moment when the cars meet is 10 s.

c) How far from the starting position A does the encounter take place?

To determine the distance from the point A where the encounter occurs, just substitute the time value in the space equation of the car A.

SA = 12.t
SA = 12.10
SA = 120 m

So we have that the cars are 120 m from the initial position A

3) (ESPM-SP) Two colleagues are in a circular square whose diameter is 60 m, according to the diagram.

The two decide to walk, starting from A, and one of them follows the diameter AC while the other follows the contour ABC. If the two walk at a speed of 1.0 m/s, the one who follows the contour passing through point B arrives at point C with a delay in relation to the colleague. Considering PI = 3.1, this delay, in seconds, is equal to:
a) 27
b) 33
c) 67
d) 73
e) 93
Solution:
Finding the time function of the space of the one who walks along the path ABC:

SABC = S0 + vA.t

The distance covered by this person is equal to half the length of the circumference.

Since the circumference of the circle is given by C = 2πR, half of that length is:

SABC = πR

So:

πR = 0 + 1.t1
3,1.30 = t1
t1 = 93 s

The time taken for the one who travels the diameter of 60 m is from:

SAC = S0 + v.t2
60 = 0 + 1.t2
t2 = 60 s.

Therefore, the difference between the times is:

t1 – t2 = 93 – 60
t1 – t2 = 33 s

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