Problem 2 The position P of a particle moving in an xy plane

The position P of a particle moving in an xy plane is given by F-(3.0-s.ooeyi+(7.00-withF in meters and t t in seconds. In unit-vector notation, calculate the following for t 2.10 s. F-1 149.587 (a) m (b) 298 × m/s (d) What is the angle between the positive direction of the x axis and a line ta ngent to the particle\'s path at t-2.a sucounterciockwoefrom th. .rais) 276 677 GO Tutorial Addtional Materials

Solution

the position vector is,

    r = [3t^3 - 5t]i^ + [7 - 8t^4] j^

at t = 2.10 s, the position is,

    r = [3(2.1)^3 - 5(2.1)]i^ + [7 - 8(2.1)^4] j^)

      = [17.283 i^ -148.5848 j^] m           ........ in vector notation

magnitude: r = sqrt[17.283*17.283 + 148.5848*148.5848] = 149.58 m

(b)

the velocity is,

    v = dr/dt

       = d/dt{[3t^3 - 5t]i^ + [7 - 8t^4] j^}

       = [9t^2 - 5]i^ + [ - 32t^3] j^

at t = 2.10 s, the velocity is,

v = [9*2.1^2 - 5]i^ + [ - 32*2.1^3] j^

    = [34.69 i^ -296.352 j^] m/s    ........in vector notation

magnitude: v = sqrt[34.69*34.69 + 296.352*296.352] = 298.37 m/s

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the acceleration is,

   a = dv/dt

      =ddt{[9t^2 - 5]i^ + [ - 32t^3] j^}

       = 18t i^ - 96t^2

at t = 2.10 s, the acceleration is,

   a = 18*2.10 i^ - 96*2.1^2

      = [37.8 i^ - 423.36 j^] m/s^2          ....... in vector notation

magnitude: a = 425 m/s^2

(d)

angle = 360 - tan^-1[-423.36/37.8] = 276^o