The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2sin πt + 5cos πt, where t is measured in seconds. (Round your answers to two decimal places.)[1,2][1,1.1][1,1.01][1,1.001]

Question

contestada

Respuesta :

nuuk nuuk · Answer 1 · 2018-09-13T14:34:14+02:00

solution:

s( t ) = ( 2 )sin( πt ) + ( 2 )cos(πt )

v( t ) = s'( t ) = ( 2π )cos( πt) - ( 2π )sin( πt )

vavg = 1 / ( b - a ) Integral a to b [ v( t ) ] dt

( a )

vavg

= 1 / ( 2 - 1 ) Integral 1 to 2 [ ( 2π)cos( πt ) - ( 2π )sin( πt ) ] dt

= [ ( 2)sin( πt ) + ( 2 )cos( πt ) ] 1to 2

= [ ( 2)sin( 2π ) + ( 2 )cos( 2π ) ] - [ ( 2 )sin( π ) + ( 2 )cos( π) ]

= 4 cm / s

( b )

vavg

= 1 / ( 1.1 - 1 ) Integral 1 to 1.1 [ ( 2π)cos( πt ) - ( 2π )sin( πt ) ] dt

= 10 [ ( 2 )sin( πt ) + ( 2 )cos( πt ) ] 1 to 1.1

= 10 [ [ ( 2 )sin( 1.1π ) + ( 2)cos( 1.1π ) ] - [ ( 2 )sin( π) + ( 2 )cos( π ) ] ]

-5.20 cm /s

( c )

vavg

= 1 / ( 1.01 - 1 ) Integral 1 to 1.01 [ ( 2π)cos( πt ) - ( 2π )sin( πt ) ] dt

= 100 [ ( 2 )sin( πt ) + ( 2 )cos( πt ) ] 1 to 1.01

= 100 [ [ ( 2 )sin( 1.01π ) + (2 )cos( 1.01π ) ] - [ ( 2 )sin( π) + ( 2 )cos( π ) ] ]

-6.18 cm /s

( d )

vavg

= 1 / (1.001 - 1 ) Integral 1 to 1.001 [ ( 2π )cos( πt ) - ( 2π )sin(πt ) ] dt

= 1000 [ ( 2 )sin( πt ) + ( 2 )cos( πt ) ] 1 to 1.001

= 1000 [ [ ( 2 )sin( 1.001π ) + (2 )cos( 1.001π ) ] - [ ( 2 )sin(π ) + ( 2 )cos( π ) ] ]

-6.27 cm /s