A particle starts to move in a straight line from a point with velocity 10 m/s and acceleration – 20 m/s²? Find the position and veloci

Question

A particle starts to move in a straight line from
a point with velocity 10 m/s and acceleration – 20 m/s²? Find the position and velocity of the
particle at (i) t = 5s, (ii) t’ = 10 s.​

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Minh Khuê 1 year 2021-08-21T23:58:54+00:00 1 Answers 8 views 0

Answers ( )

    0
    2021-08-22T00:00:36+00:00

    Answer: s(5) = -200,  v(5) = -90

                  s(10) = -900, v(10) = -190

    Explanation:

    Position: s(t)

    Velocity:  s'(t) = v(t)                ⇒      s(t)=\int {v(t)} \, dt

    Acceleration     v'(t) = a(t)      ⇒      v(t)=\int {a(t)} \, dt

    We are given that acceleration a(t) = -20   and velocity v(t) = 10

    v(t)=\int {a(t)} \, dt\\\\v(t)=\int{-20}\, dt\\\\v(t)=-20t + C \\\\v(t)=10\quad \longrightarrow \quad C=10\\\\v(t)=-20t+10

    s(t)=\int {v(t)} \, dt\\\\s(t)=\int {(-20t+10)} \, dt\\\\s(t)=-10t^2+10t\\\\

    (a) Input t = 5 into the s(t) and v(t) equations

       s(5) = -10(5)² + 10(5)           v(5) = -20(5) + 10

              =  -250  +  50                     = -100   +  10

              =       -200                          =        -90

    (b) Input t = 10 into the s(t) and v(t) equations

         s(10) = -10(10)² + 10(10)           v(10) = -20(10) + 10

                =  -1000  +  100                     = -200   +  10

                =       -900                          =        -190

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