College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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please solve ; 4.4
4. A particle has a position changing with time as per the equations below:
x = 0.28t + 0.036t2
y = 0.019t
Calculate the following:
4.1 The particle's displacement between t=0s and t=2s.
4.2 The particle's average velocity between t=0s and t=3s.
4.3 The particle's average acceleration between t=1s and t=3s.
4.4 The particle's instantaneous acceleration's magnitude and direction between t=1s and t=3s.
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Transcribed Image Text:4. A particle has a position changing with time as per the equations below: x = 0.28t + 0.036t2 y = 0.019t Calculate the following: 4.1 The particle's displacement between t=0s and t=2s. 4.2 The particle's average velocity between t=0s and t=3s. 4.3 The particle's average acceleration between t=1s and t=3s. 4.4 The particle's instantaneous acceleration's magnitude and direction between t=1s and t=3s.
Given: x = 0.28t +0.036t? and
y = 0.0191
Let i and jbeunit vectors along x and y directions.
4.1
Displacement d is change in position.
At,t = 0,x, = 0 and
Y, = 0
At,t = 2s,x, = (0.28×2)+0.036×2² = 0.704m and
y, = 0.019×2' = 0.152m
= (x, -x,)î +(y, - y,)i putting values
d = (0.704 m)î + (0.152m) j
[Answer]
y, = 0
Att = 3s,x, = 0.28×3+0.036×3² = 1.164m and y, = 0.019x3 = 0.513m
4.2
Att = 0,x, = 0
and
d = (x, -x,)î+(y, - y,)} putting values d=1.164mî +0.513m j
At = 3 s
Average velocity( Vang ) is displacement per unit time.
1.164 mi +0.513m j
V
%3D
%3D
ang
ang
At
3s
V = (0.388m/s)i +(0.171m/s)}
[Answer]
ang
dx
=v, = 0.28 +0.072t and v
dt
dy
→v, = 0.0571?
dt
4.3
V, =
At t = ls,v, = 0.28+0.072×1= 0.352 m/s and
v = 0.057×1² = 0.057m/ s
Att = 3s ,v = 0.28+0.072×3= 0.496m/s and
V =0.057×3² = 0.513m/s
At = 2s
Average acceleration(a) is changein velocity per unit time.
(v, –v,)î+(v, -v,)
Δν
putting values
At
At
[(0.496 – 0.352)m /s]i+[(0.513-0.057)m/s]j
Ang
2.s
(0.072 m/ s²)î+(0.228 m/ s³)j
[Answer]
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Transcribed Image Text:Given: x = 0.28t +0.036t? and y = 0.0191 Let i and jbeunit vectors along x and y directions. 4.1 Displacement d is change in position. At,t = 0,x, = 0 and Y, = 0 At,t = 2s,x, = (0.28×2)+0.036×2² = 0.704m and y, = 0.019×2' = 0.152m = (x, -x,)î +(y, - y,)i putting values d = (0.704 m)î + (0.152m) j [Answer] y, = 0 Att = 3s,x, = 0.28×3+0.036×3² = 1.164m and y, = 0.019x3 = 0.513m 4.2 Att = 0,x, = 0 and d = (x, -x,)î+(y, - y,)} putting values d=1.164mî +0.513m j At = 3 s Average velocity( Vang ) is displacement per unit time. 1.164 mi +0.513m j V %3D %3D ang ang At 3s V = (0.388m/s)i +(0.171m/s)} [Answer] ang dx =v, = 0.28 +0.072t and v dt dy →v, = 0.0571? dt 4.3 V, = At t = ls,v, = 0.28+0.072×1= 0.352 m/s and v = 0.057×1² = 0.057m/ s Att = 3s ,v = 0.28+0.072×3= 0.496m/s and V =0.057×3² = 0.513m/s At = 2s Average acceleration(a) is changein velocity per unit time. (v, –v,)î+(v, -v,) Δν putting values At At [(0.496 – 0.352)m /s]i+[(0.513-0.057)m/s]j Ang 2.s (0.072 m/ s²)î+(0.228 m/ s³)j [Answer]
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