2.68 CALC An object's velocity is measured to be U₁(1) = a - Br², wher

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2021-10-28

2.68 CALC An object's velocity is measured to be U₁(1) = a - Br², where a = 4.00 m/s and ß = 2.00 m/s³. At t = 0 the object is at x = 0. (a) Calculate the object's position and accel cration as functions of time. (b) What is the object's maximum positive displacement from the origin?

Answer & Explanation

Jeffrey Jordon

Jeffrey Jordon

Expert2021-11-02Added 2605 answers

Given data:

The velocity of the particle is

vx(t)=αβt2

Where

α=4 ms1

β=2 ms1

(a). The object's position from the velocity is defined as,

dxdt=vx(t)

dx=vx(t)dt

Substituting the value of velocity,

dx=(αβt2)dt

Integrating with respect to time,

x=0t(αβt2)dt

x=[αtβt33]0t

x=αtβt33

Substituting the given values,

x=4t2t33

This equation describes the position in terms of time.

(b). The acceleration in terms of time is,

ax(t)=dvx(t)dt

ax(t)=ddt(αβt2)

ax(t)=2βt

Substituting the given value,

ax(t)=2×2×t

ax(t)=4t

This expression describes the acceleration in terms of time.

madeleinejames20

madeleinejames20

Skilled2023-05-19Added 165 answers

(a) To find the object's position and acceleration as functions of time, we'll integrate the given expression for velocity.
Given:
U1(1)=aBr2
a=4.00m/s and β=2.00m/s3
To find the object's position, we integrate the velocity function with respect to time:
U1(1)dt=(aBr2)dt
Integrating both sides:
x(t)=(aBr2)dt
Integrating term by term:
x(t)=adt(Br2)dt
Simplifying the integrals:
x(t)=at(Br2)dt
Now, we need to find the integral of Br2 with respect to time. Since we have the expression for velocity in terms of position (U1(1)), we can relate r to x using the chain rule:
r=x
Differentiating with respect to time, we get:
drdt=12x·dxdt
dxdt=2x·drdt
Substituting this back into the integral for x(t):
x(t)=atB(2x)2dt
x(t)=at4Bxdt
Integrating both sides:
x(t)=at4Bxdt
Simplifying the integral:
x(t)=at4Bxdt
x(t)=at4B·12x2+C
Here, C is the constant of integration. Since the object is at x=0 when t=0, we can substitute these values to find the specific solution for C:
x(0)=0
a·04B·12·02+C=0
C=0
Therefore, the object's position as a function of time is:
x(t)=at4B·12x2
To find the object's acceleration, we differentiate the velocity function with respect to time:
dU1(1)dt=ddt(aBr2)
dU1(1)dt=2Br·drdt
dU1(1)dt=2Br·12x·dxdt
dU1(1)dt=B·rx·dxdt
Substituting dxdt=2x·drdt:
dU1(1)dt=<br>B·rx·2x·drdt
dU1(1)dt=2Br·drdt
dU1(1)dt=2Br·drdt
Now, we can substitute r=x and drdt=12x·dxdt:
dU1(1)dt=2Bx·12x·dxdt
dU1(1)dt=B·dxdt
Therefore, the object's acceleration as a function of time is:
a(t)=B·dxdt
(b) To find the object's maximum positive displacement from the origin, we need to find the maximum value of x(t). Since the coefficient of x2 in the position function is negative, the maximum value of x(t) occurs at the vertex of the parabolic curve.
The vertex of a parabola with the equation ax2+bx+c is given by:
xvertex=b2a
In our case, a=2B and b=a. Substituting these values, we have:
xvertex=a2(2B)=a4B
Since a=4.00m/s and B=2.00m/s3, we can substitute these values to find the maximum positive displacement:
xvertex=4.004·2.00=4.008.00=0.50m
Therefore, the object's maximum positive displacement from the origin is 0.50m.
Nick Camelot

Nick Camelot

Skilled2023-05-19Added 164 answers

Answer:
a)A(t)=2Bt
b) 11.43m
Explanation:
(a) Position and acceleration as functions of time:
The object's velocity, U1(t), is given by U1(t)=aBr2, where a=4.00m/s and B=2.00m/s3.
To find the position function, we need to integrate the velocity function with respect to time:
X(t)=U1(t)dt=(aBr2)dt
Integrating with respect to time, we get:
X(t)=atB3t3+C1
where C1 is the constant of integration. Since the object is at x=0 when t=0, we can substitute these values into the equation to solve for C1:
0=a·0B3·03+C1
Simplifying, we find that C1=0. Therefore, the position function is:
X(t)=atB3t3
To find the acceleration function, we need to take the derivative of the velocity function with respect to time:
A(t)=dU1(t)dt=ddt(aBr2)
Differentiating with respect to time, we get:
A(t)=02Bt=2Bt
So the acceleration function is:
A(t)=2Bt
(b) Maximum positive displacement from the origin:
The maximum positive displacement from the origin occurs when the velocity is zero. So we need to find the time when U1(t)=0:
0=aBr2
Solving for t, we get:
t=aB
Substituting the given values of a and B, we have:
t=4.00m/s2.00m/s3=2s2=2s
Finally, substituting this value of t into the position function, we can find the maximum positive displacement:
X(2s)=a(2s)B3(2s)3
Simplifying, we get:
X(2s)=4.00m/s·2s2.00m/s33·(2s)3
Calculating this expression will give us the maximum positive displacement from the origin.
X(2s)=4.00m/s·2s2.00m/s33·(2s)3
Simplifying further:
X(2s)=4.00m/s·2s2.00m/s33·2s3
X(2s)=4.00m/s·2s2.00m/s33·8s3
X(2s)=4.00m/s·2s16.00m/s33·s3
Calculating the values:
X(2s)5.66m17.09m=11.43m
Therefore, the object's maximum positive displacement from the origin is approximately 11.43m in the negative direction.
Eliza Beth13

Eliza Beth13

Skilled2023-05-19Added 130 answers

Step 1:
(a) To calculate the object's position and acceleration as functions of time, we need to integrate the given velocity function.
Given: U1(1)=aBr2, where a=4.00m/s and ß=2.00m/s3.
To find the object's position, we integrate the velocity function with respect to time:
x(t)=U1(t)dt
x(t)=(aBr2)dt
Integrating with respect to t gives:
x(t)=adtBr2dt
x(t)=atB3t3+C
where C is the constant of integration.
To find the object's acceleration, we differentiate the velocity function with respect to time:
a(t)=dU1(t)dt
a(t)=ddt(aBr2)
a(t)=ddt(aBx2)
Differentiating with respect to t gives:
a(t)=ddt(a)ddt(Bx2)
a(t)=0ddt(Bx2)
a(t)=2Bxdxdt
Substituting the expression for dx/dt obtained earlier:
a(t)=2Bx(ddt(atB3t3+C))
a(t)=2Bx(aBt2)
a(t)=2Ba+2B2t2x
Step 2:
(b) To find the object's maximum positive displacement from the origin, we need to determine when the velocity function changes from positive to negative. This occurs at the turning point, where the velocity is zero.
Setting U1(t)=0, we have:
aBr2=0
Br2=a
r2=aB
r=aB
To find the object's maximum positive displacement from the origin, we substitute r=aB into the position function x(t):
x(t)=atB3t3+C
Substituting r=aB gives:
x(t)=atB3t3+C
x(t)=atB3t3+C
x(t)=atB3t3+C
To find the constant C, we use the initial condition given that at t=0, the object is at x=0. Substituting these values into the position function:
x(0)=a(0)B3(0)3+C
0=0+0+C
C=0
Therefore, the position function becomes:
x(t)=atB3t3
To find the maximum positive displacement, we need to find the maximum of the position function. Taking the derivative of x(t) with respect to t:
dxdt=aBt2
Setting dxdt=0 to find the turning point:
aBt2=0
Bt2=a
t2=aB
t=aB
Substituting this value of t back into the position function:
x(aB)=a(aB)B3(aB)3
x(aB)=aaBB3(aB)3
x(aB)=aaBB3(aB)aB
x(aB)=aaBa3aB
x(aB)=2a3aB
Therefore, the object's maximum positive displacement from the origin is 2a3aB.

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