Question

Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data v(t)=sin(t)−cos(t),s(0)=−10. Find a function describing the position of the particle.

Answer 1

Answer:

s(t) = -cos(t) -sin(t) -9

Step-by-step explanation:

The velocity (v) of a moving particle is the rate of change of poisition (s)  i.e

$v=\frac{ds}{dt}$

from the above equation s becomes:

$s=\int\(v(t)} \, dt$

from the given data

$v(t) = sin(t) - cos(t)$

therefore,$s(t) = \int\ {sin(t)- cos(t)} \, dt\\ s(t) = \int\ sin(t) dt - \int\ cos(t) dt\\\\s(t) = -cos(t) - sin(t) +C\\put \\t=0\\s(0) = -cos(0) - sin(0) +C\\-10 = -1 -0 +C\\C = -10+1\\C=-9\\therfore,\\s(t) = -cos(t) -sin(t) -9$

Answer 2

The function describing the position of the particle is  S= -cost-sint -9.

Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. T

The particle is moving according to the data v(t)=sin(t)−cos(t),s(0)=−10.

Integration;

Integration is the reverse of differentiation.

The velocity (v) of a moving particle is the rate of change of position (s) is given b;

$\rm V=\dfrac{ds}{dt}$

To find a function describing the position of the particle integrating the function on both sides.

Therefore,

The position of the particle is;

$\rm S=\int\limits v(t)dt$

Substitute the value of v(t) in the equation

$\rm S=\int\limits v(t)dt\\\\\rm S=\int\limits (sin(t)-cos(t))dt\\\\S=\int\limits sintdt-\int\limits costdt\\\\S=-cost-sint+c\\\\When \ t-0\ and\ s(0)=-10\\\\-10 = -cos(0)-sin(0)+c\\\\c=-10+1\\\\c=-9\\\\S=-cost-sint-9$

Hence, the function describing the position of the particle is  S= -cost-sint -9.

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