velocity and acceleration

Question

Answer 1

a)

Position vector,r = (2t^3/2,-3t , 6t^3/2)

i) Calculate the position vector at time t = 9 sec:

r = (2t^3/2,-3t , 6t^3/2)

   = (2(9)^3/2,-3(9), 6(9)^3/2)

   = (2(3²)^3/2,-3(9), 6(3²)^3/2)

   = (2(27),-3(9), 6(27))

   = (54,-27, 162)

ii) Calculate the velocity vector at time t = 9 sec:

v = r' = (d/dt(2t^3/2),d/dt(-3t) ,d/dt(6t^3/2))

            = ((3/2)(2t^3/2-1),-3 ,(3/2)(6t^3/2-1))

             = (3t^1/2,-3 ,9t^1/2)   

Substitute 9 for t.                      

 v = (3(9)^1/2,-3 ,9(9)^1/2)

    = (3(3²)^1/2,-3, 9(3²)^1/2)

   = (3(3),-3, 9(3))

   = (9,-3, 27)

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iii) Calculate the accelaration vector at time t = 9 sec:

a =v' = d/dt(3t^1/2,-3 ,9t^1/2)

            = (3d/dt(t^1/2),d/dt(-3) ,9d/dt(t^1/2))

            = ((1/2)(3t^1/2-1),0 ,(1/2)(9t^1/2-1))

             = ((3/2)t^-1/2,0 ,(9/2)t^-1/2)   

Substitute 9 for t.                      

 a = ((3/2)(9)^-1/2,0 ,(9/2)(9)^-1/2)

    = ((3/2)(3²)^-1/2,0, (9/2)(3²)^-1/2)

   = (1/2,0, 3/2)

 

 

Answer 2

b)

v = (3t^1/2,-3 ,9t^1/2)  

Calculate the magnitude:

v =√((3t^1/2)^2+(-3)^2+(9t^1/2)^2)

   =√9t+9+81t)

    =√(90t+9)

   √(90t+9) =8

Square on both sides

90t+9 =64

90t =55

t =0.611

c)

Scaler product of velocity of acceleration is 0 when velocity and acceleration are perpendicular.

 

Comments

Isn't c) it does not vary with time as a different value of t will still produce the same dot product?