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The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try?
Most Upvoted Answer
The acceleration of a particle is defined by the relation a=0.4(1-kv),...
Answer:
Given:
a = 0.4(1-kv)
At t=0, u=0, x=4m
At t=15s, v=4m/s
(a) Determine the constant k:
a = 0.4(1-kv)
At t=0, u=0, x=4m
a = 0.4(1-k(0))
a = 0.4
At t=15s, v=4m/s
a = 0.4(1-kv)
0.4 = 0.4(1-k(4))
0.4 = 0.4(1-4k)
1-4k = 1
-4k = 0
k = 0
Therefore, k = 0.
(b) The position of the particle when v=6m/s:
We need to find the position x when v=6m/s.
Given, a = 0.4(1-kv)
At t=0, u=0, x=4m
Let's integrate the expression to find the velocity function:
a = dv/dt
0.4(1-kv) = dv/dt
0.4dt = dv/(1-kv)
∫0 to t 0.4dt = ∫0 to v dv/(1-kv)
0.4t = (-1/k)ln(1-kv) + C
At t=0, u=0, x=4m
0 = (-1/k)ln(1-k(0)) + C
C = 0
0.4t = (-1/k)ln(1-kv)
When v=6m/s,
0.4t = (-1/k)ln(1-k(6))
0.4t = (-1/0)ln(1)
t = 0
We got t=0, which means the particle hasn't moved when v=6m/s. Therefore, the position of the particle when v=6m/s is 4m.
(c) The maximum velocity of the particle:
The maximum velocity occurs when the acceleration is zero, i.e., when a=0.
Given, a = 0.4(1-kv)
0 = 0.4(1-kv)
1-kv = 0
v = 1/k
Substituting k=0, we get v=∞.
Therefore, the maximum velocity of the particle is infinity.
Given:
a = 0.4(1-kv)
At t=0, u=0, x=4m
At t=15s, v=4m/s
(a) Determine the constant k:
a = 0.4(1-kv)
At t=0, u=0, x=4m
a = 0.4(1-k(0))
a = 0.4
At t=15s, v=4m/s
a = 0.4(1-kv)
0.4 = 0.4(1-k(4))
0.4 = 0.4(1-4k)
1-4k = 1
-4k = 0
k = 0
Therefore, k = 0.
(b) The position of the particle when v=6m/s:
We need to find the position x when v=6m/s.
Given, a = 0.4(1-kv)
At t=0, u=0, x=4m
Let's integrate the expression to find the velocity function:
a = dv/dt
0.4(1-kv) = dv/dt
0.4dt = dv/(1-kv)
∫0 to t 0.4dt = ∫0 to v dv/(1-kv)
0.4t = (-1/k)ln(1-kv) + C
At t=0, u=0, x=4m
0 = (-1/k)ln(1-k(0)) + C
C = 0
0.4t = (-1/k)ln(1-kv)
When v=6m/s,
0.4t = (-1/k)ln(1-k(6))
0.4t = (-1/0)ln(1)
t = 0
We got t=0, which means the particle hasn't moved when v=6m/s. Therefore, the position of the particle when v=6m/s is 4m.
(c) The maximum velocity of the particle:
The maximum velocity occurs when the acceleration is zero, i.e., when a=0.
Given, a = 0.4(1-kv)
0 = 0.4(1-kv)
1-kv = 0
v = 1/k
Substituting k=0, we get v=∞.
Therefore, the maximum velocity of the particle is infinity.
Community Answer
The acceleration of a particle is defined by the relation a=0.4(1-kv),...
Since, a=dv/dt integrate the given relation with respect to time and apply the boundary conditions to get the value of "k".
Since, a=vdv/dx integrate the the relation with respect to displacement and apply the boundary conditions to get the value of "x"
for maximum velocity a=0 and find the Vmax from the original given equation.
Since, a=vdv/dx integrate the the relation with respect to displacement and apply the boundary conditions to get the value of "x"
for maximum velocity a=0 and find the Vmax from the original given equation.
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The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try?
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The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try?.
The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The acceleration of a particle is defined by the relation a=0.4(1-kv), where k is a constant. Knowing that t=0 the particle starts from rest at x=4m and that when t=15s v=4m/s (a)determine the constant k( b) the position of the particle when v=6m/s c) the maximum velocity of the particle. Any try?.
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