Attempt Results Entered Answer Previeww -180 -180 At least one of the answers ab

Question

Attempt Results Entered Answer Previeww -180 -180 At least one of the answers above is NOT correct. 1 of the questions remains unanswered. (1 point) A particle has zero velocity initially (i.e., at time t = 0 ) and its acceleration at second. During the time interval [3, 6] seconds, seconds is a(t) = 50t-4t3 meters-per-second per () the average acceleration of this particle is-180 (i) the average speed of this particle is Hint meters-per-second per second meters-per-second Note: You can earn partial credit on this problem.

Explanation / Answer

1)average acceleration =[a(6)-a(3)]/(6-3)

=[50(6)-4(6)3   -50(3)+4(3)3]/(6-3)

=-606/3

=-202 m/s/s

2)a(t)=50t -4t3

velocity v(t)=integral a(t)

=integral 50t -4t3 dt=(50/2)t2 -(4/4)t4+c =25t2 -t4+c

given initial velocity =0

==>v(0)=0

v(0)=25*02 -04+c=0 ==>c=0

v(t)=25t2 -t4

average speed=[v(6)-v(3)]/[6-3]

=[25*62 -64   -25*32 +34]/[3]

=-540/3

=-180m/s

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