Question

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion

s = 4 sin(πt) + 2 cos(πt),

where t is measured in seconds. (Round your answers to two decimal places.)

Answer 1

Complete Question

The displacement (in centimetres) of a particle moving back and forth along a straight line is given by the equation of motion

s = 4 sin(πt) + 2 cos(πt),

where t is measured in seconds. (Round your answers to two decimal places.)

(a) Find the average velocity during each time period.

i. [1,2]

ii. [1 ,1.1]

iii. [1, 1.01]

Answer:

$V_A_1=4m/s$

$V_A_2=-11.38m/s$

$V_A_3=-12.47m/s$

Step-by-step explanation:

From the question we are told that

Equation of motion is given by

$s = 4 sin(\pi t) + 2 cos(\pi t)$

Generally equation for average velocity is mathematically given by

  $V_A=\frac{S(y)-S(x)}{y-x}$

  $V_A=\frac{4 sin(\pi t) + 2 cos(\pi t)(y)-(4 sin(\pi t) + 2 cos(\pi t))(x)}{y-x}$

a) Co-ordinate(1,2)

 $V_A=\frac{4 sin(\pi y) + 2 cos(\pi y)-(4 sin(\pi x) + 2 cos(\pi x))}{y-x}$

 $V_A=\frac{4 sin(\pi 2) + 2 cos(\pi 2)-(4 sin(\pi) + 2 cos(\pi))}{2-1}$

 $V_A=\frac{0+ 2*1-(0 + 2)*-1}{2-1}$

 $V_A_1=4m/s$

b)Co-ordinate(1,1.1)

    $V_A_2=\frac{4 sin(\pi 1.1) + 2 cos(\pi 1.1)-( sin(\pi ) + 2 cos(\pi ))}{1.1-1}$

    $V_A_2=\frac{0+ 2*-1-(-0.30902*2)+2*-0.95106}{1.1-1}$

    $V_A_2=\frac{((-0.30902*2)+2*-0.95106)-(0+ 2*-1)}{1.1-1}$

    $V_A_2=\frac{((-3.1382)-(-2)}{1.1-1}$

    $V_A_2=-11.38m/s$

c) Co-ordinate(1,1.01)

    $V_A_3=\frac{4 sin(\pi 1.01) + 2 cos(\pi 1.01)-( sin(\pi ) + 2 cos(\pi)}{1.01-1}$

    $V_A_3=\frac{0.2214045768+1.996933901)-(-2)}{1.01-1}$

    $V_A_3=\frac{(-2.12)-(-2)}{1.01-1}$

    $V_A_3=-12.47m/s$