A particle moves along the x-axis with an acceleration givenby: a(t) = 5sin(wt)

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 displacement of the particle between t = 0 secand t = 4 sec.

PLEASE SHOW ANDEXPLAIN ALL STEPS! PLEASE SHOW ANDEXPLAIN ALL STEPS!

Explanation / Answer

a(t) = 5sin(0.7t) ft/sec2   v(t) = a(t) dt = -(5/0.7)cos(0.7t) + C = -7.14 cos(0.7t) + C we know v(0) = 5 ft/s so 5 = -7.14 + C C = 12.14 ft/s v(t) = -7.14 cos(0.7t) + 12.14 ft/s x(t) = v(t) dt =-7.14/0.7 sin(0.7t) + 12.14t + C' = -10.2 sin(0.7t) + 12.14t +C' we know x(0) = 2 ft so C' = 2 ft x(t) = -10.2 sin(0.7t) + 12.14t + 2 ft displacement = x(4) - x(0) = -10.2 sin(0.7*4) + 12.14*4 = 45.1ft This question only gives the speed = 5 ft/s at t =0, it may be v(0) = +5 ft/s or -5 ft/s This question only gives it is 2 ft away from theorigin at t = 0, it may be x(0) = +2 ft or -2 ft

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