1.3(t 5.0)] m/s. This decrease A particle at rest leaves the origin with its vel
ID: 1576290 • Letter: 1
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 tExplanation / 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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.