A particle moves along the x-axis with an acceleration givenby: a(t) = 5sin(wt)
ID: 1759536 • Letter: A
Question
A particle moves along the x-axis with an acceleration givenby:a(t) = 5sin(wt) ft/sec2 (where "w" =0.7rad/sec)
Initially (@ t = 0) the particle is 2ft away from the origin and ismoving with a speed of:
v(t) = 5 ft/sec
(a) Determine the velocity and position of the particle asfunctions of time.
[Part (a) has been posted as two separatequestions...once you know part (a), you can find part (b)below]
(b) Determine the distance traveled by the particle between t = 0sec and t = 4 sec.
PLEASE SHOW ANDEXPLAIN ALL STEPS! PLEASE SHOW ANDEXPLAIN ALL STEPS!
Explanation / Answer
= 5 [-cos wt ] / w + C = ( -5 / w ) cos wt +C -----( 1) where C = integration constant According to problem at t= 0 , speed v( t ) = 5 ft/ s substitue this condition in eq ( 1 ) we get 5 = ( -5 / 0.7 ) cosw ( 0 ) + C 1 = -( 1/ 0.7 ) + C C = 2.428 So, velocity of the particle as a function of time is v(t ) = ( -5 / w ) cos wt + 2.428 = -7.142 cos wt + 2.428 since w = 0.7 rad / sInitially (@ t = 0) the particle is 2ft away from the origin and ismoving with a speed of:
v(t) = 5 ft/sec
( b ) . position x (t) = integral v( t ) dt = -7.142[sin wt / w ] + 2.428 t + C ' = -10.204 sin wt + 2.428 t + C ' ----(2) where C ' = integration constant At t= 0 , x = 2 substitue this condition in eq ( 2 ) we get 2 = -10.204 sin w( 0 ) + 2.428( 0 ) + C ' C ' = 2 So, function of position x( t ) = -10.204 sin wt + 2.428t + 2 Therefore the distance traveled by the particle between t = 0sec and t = 4 sec is X = x( 4 ) - x( 0 ) plug the values we get X = -10.204 sin [0.7 * 4 ] +2.428* 4 +2 - [ -10.204 sin(0.7*0)-2.428*0 +2] = 8.2937-2 = 6.2937
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