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3 years ago

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A particle moves along the x-axis with velocity v(t) = t2 - 4, with t measured in seconds and v(t) measured in feet per second.

Find the total distance travelled by the particle from t = 0 to t = 3 seconds. (4 points)

1 answer:

Nadya [2.5K]3 years ago

8 0

Answer:

Total distance D=7.66 ft

Explanation:

Given that: $V=t^{2}-4$

Here velocity of particle will be zero at t=2 sec,so we take interval to find total distance.

We know that $V=\dfrac{ds}{dt}$

ds=V dt

So $s=\int_{t_{1}}^{t_{2}}Vdt$

It means that ,the area of velocity-time(V-t) graph will give the displacement.

Given that $t_{1}=0 ,t_{2}=3$

So now by putting the value in above integration

$s=\int_{0}^{2}(t^{2}-4)dt+\int_{2}^{3}(t^{2}-4)$

$s=\left [\frac{1}{3}t^3-4t\right ]_o^2+\left [\frac{1}{3}t^3-4t\right]_2^3$

s= -5.33+2.33 ft

s= -3 ft (we know that displacement is a vector quantity so it have sing)

So this is the displacement of particle at time 0 sec to 3 sec.

To find the total distance we will add all take mode on -5.33 ft and and will add with 2.33 ft instead of subtract.

So the total distance travelled by particle D=7.66 ft.

D=7.66 ft

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a)

The momentum of an object is a vector quantity given by:

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v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

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$p=m_1 u_1 + m_2 u_2$

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The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

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$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

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$m_1 = 30 kg$ is the mass of the girl

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We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

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$\Delta p$ is the change in momentum

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