1.3(t 5.0)] m/s. This decrease A particle at rest leaves the origin with its vel

Question

1.3(t 5.0)] m/s. This decrease A particle at rest leaves the origin with its velocity increasing with time according to v(t)-3.9t m/s. At 5.0 s, the particle's velocity starts decreasing according to [19.5 continues until t11.0 s, atter which the particle's velocity remains constant at 11.7 m/s (a) What is the acceleration of the particle as a function of time? (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume that a is in m/s and t is in seconds. Do not include units in your answer.) 1.9 t

Explanation / Answer

for 0<t<5:

velocity=v=3.9*t

==>position=x=integration of v*dt

=integration of 3.9*t*dt

=1.95*t^2+c

where c is a constant of integration.

at t=0, x=0

==>c=0

at t=5 , x=1.95*5^2=48.75 m

for 5<t<11:

v=19.5-1.3*(t-5)=-1.3*t+26

position x=integration of v*dt

=-0.65*t^2+26*t+c

where c is constant of integration

at t=5, x=48.75 m

==>-0.65*5^2+26*5+c=48.75

==>c=-65 m

at t=11, x=-0.65*11^2+26*11-65=142.35

for t>11:

v=11.7

==>x=integration of v*dt

=11.7*t+c

where c is constant of integration

at t=11, x=142.35 m

==>11.7*11+c=142.35

==>c=13.65

==========================================

part b:

at t=2 seconds, x=1.95*2^2=7.8 m

at t=7 seconds, x=-0.65*7^2+26*7-65=85.15 m

at t=12 seconds, x=11.7*12+13.65=154.05 m

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