Direction: Solve the problem. X2- 5000 m X- Om t- Os ts- 5s ta- 10s X =? A x = R1) = t -t +2 a =? B - 3r* +++t 1. Sub-atomic particles A and B are moving in a 100-meter dash line in 10s. Particle A moves such that its displacement x in meters is given by the function x = (t -t + 2)meters. Meanwhile, particle B is moving at a velocity v = (3t' + -+t) mvs. If both particles A and B started from rest at the same time, Compute the following; Average velocity v of purticle A and purticle B in m/s in 10s trip. b. Displacement x of particle A and particle B from the starting point at t= 5s. c. Velocity v of purticle A and particle B at t 5s a.
Direction: Solve the problem. X2- 5000 m X- Om t- Os ts- 5s ta- 10s X =? A x = R1) = t -t +2 a =? B - 3r* +++t 1. Sub-atomic particles A and B are moving in a 100-meter dash line in 10s. Particle A moves such that its displacement x in meters is given by the function x = (t -t + 2)meters. Meanwhile, particle B is moving at a velocity v = (3t' + -+t) mvs. If both particles A and B started from rest at the same time, Compute the following; Average velocity v of purticle A and purticle B in m/s in 10s trip. b. Displacement x of particle A and particle B from the starting point at t= 5s. c. Velocity v of purticle A and particle B at t 5s a.
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Transcribed Image Text:Direction: Solve the problem.
X2= 5000 m
X1 = Om
t = Os
ts= 5s
tz = 10s
DA = ?
ig = ?
XA = ?
A
vg =?
x = t) = t? -t + 2
aA =?
ag =?
B
v = = 3r* + ++t
1. Sub-atomic particles A and B are moving in a 100-meter dash line in 10s. Particle A moves such that its
displacement x in meters is given by the function x = (t? – t+ 2)meters. Meanwhile, particle B is moving at a
velocity v = (3r' + - t² + t) m/s. If both particles A and B started from rest at the same time, Compute the
following;
a. Average velocity v of particle A and particle B in m/s in 10s trip.
b. Displacement x of particle A and particle B from the starting point at t = 5s.
c. Velocity v of particle A and particle B at t = 5s
d. Acceleration a of particle A and particle B at t = 5s.
(Hint: wcelention is the de ri vative of velocity; ainst =
dv
dt
2. In a graphing paper, graph the function ix = (t² – t + 2) in displacement x vs time t graph (x vs t) showing
displacement x in the y-axis and time t in the x-axis. Then, graphically obtain the instantaneous velocity of
particle A at time t = 5 seconds by getting the slope of a line tangent to the curve.
Transcribed Image Text:Table
Student
No.
x = f (t)
20
f(t) = 20t³ – 19t² + t
f(t) = t³ +38t² + t
= 2t³ + 37t² +t
f(t) = 21t³ – 18t² + t
f(t) = 22t³ – 17t² + t
f(t) = 23t³ – 16t² + t
f(t) = 24t³ – 15t² + t
f(t) = 25t³ – 14t² + t
26t
1
21
2
f(t)
f(t) = 3t³ + 36t² + t
= 4t3 + 35t2 + t
21
3
23
24
25
f(t)
f(t) = 5t³ + 34t² + t
f(t)
= 7t³ + 32t² +t
4
5
6t³ + 33t² + t
26
13t? +t
f(t)
f(t) = 27t³ – 12t² + t
f(t) = 28t³ – 11t² + t
f(t) = 29t³ – 10t² + t
f(t) = 30t³ – 9t² + t
= 31t3 – 8t? +t
6.
7
f(t)
f(t) = 8t³ + 31t² +t
f(t) = 9t³ + 30t² + t
f(t) = 10t3 + 29t² + t
f(t) = 11t³ + 28t² + t
f(t) = 12t³ + 27t² + t
f(t) = 13t³ + 26t² + t
f(t) = 14t³ + 25t² + t
f(t) = 15t3 + 24t² + t
f(t) = 16t³ + 23t² + t
f(t) = 17t³ + 22t² + t
f(t) = 18t³ + 21t² +t
f(t)
27
28
29
10
30
11
31
f(t)
f(t) = 32t3 – 7t² + t
f(t) = 33t³ – 6t² +t
= 34t3 – 5t? +t
12
32
13
33
f(t)
f(t) = 35t³ – 4t² + t
f(t) = 36t3 – 3t² + t
f(t) = 37t³ – 2t² + t
f(t) = 38t³ – t² + t
14
34
35
15
16
36
17
37
18
38
19
19t3 + 20t2 +t
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