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3 years ago

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The velocity v(t) of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s2)t - (6.2 m/s3)t2. What is the ave

rage acceleration of the particle between t = 1.0 s and t = 2.0 s (up to two significant figures) ? 
A) -13 m/s2 
B) -14 m/s2
C) 13 m/s2
D) 15 m/s2
E) None of the above

1 answer:

pantera1 [17]3 years ago

6 0

Answer:

average acceleration = −14.5 m/s²

so correct option is B) -14 m/s²

Explanation:

given data

v(t) = (2.3 m/s) + (4.1 m/s2)t - (6.2 m/s3)t²

t1 = 1.0 s

t2 = 2.0 s

to find out

What is the average acceleration

solution

we know that average rate of change on the interval is

average acceleration = $\frac{Vf-Vi}{t2 - t1}$ ............1

so here put value of t for final f and initial i

we get Vf = 2.3 + (4.1)(2) - (6.2)(2)²

and Vi = 2.3 + (4.1)(1) - (6.2)(1)²

and t2 - t1 = 2 - 1 = 1 s

put value in equation 1

average acceleration = [2.3 + (4.1)(2) - (6.2)(2)² ] - [ 2.3 + (4.1)(1) - (6.2)(1)² ]

average acceleration = −14.5 m/s²

so correct option is B) -14 m/s²

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Answer:

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Explanation:

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Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

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