Problem 5
Solve.12+2x = 13+3x
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Displacement, velocity and acceleration Ex. a particle moves back and forth along a straight line. Its displacement at time t ( t is in seconds, s(t)) is in feet is s(t)=t^3 12t^2 +36t a. What is the particle’s velocity at time t v(t) = s’(t) =3t^2 24t + 36 ft/sec b. When is the particle at rest Solve v(t)= 3t^2 24t + 36 = 0 Divide by three t^2 8t + 12 = 0 factor (t 2)(t + 6) = 0 t = 2 sec, 6 sec c. what is the initial velocity Plug in 0 for t v(0) = 36 ft/sec d. what is the displacement at t =2 and t =6 sec Plug in 2 and 6 to the original function s(2) = 32 ft s(6) = 0ft Line graph Graph e. When is
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Chapter 3.4, Problem 5 is Solved
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