Question

Suppose that, at time t = 0, a particle with mass 3 has position vector ⃗r(0) = 4⃗j − ⃗k and velocity ⃗v(0) = −5⃗j − 13⃗k. The particle is then subjected to a constant force of F⃗ = 9⃗ı + 6⃗k.

(a) Find the position of the particle (as a function of time).
(b) When is the particle moving most slowly? Compare the minimum speed with the speed

at times t = 1 and t = 4.

Thank you!

My Mistake.

Answer 1

a) a = F/m = 3i + 2 k

x = 0.5 a_x t^2 = 1.5 t^2

y = 4 - V_y t = 4 - 5t

z = - 1 - 13t + 0.5*2*t^2 = -1 -13t + t^2

Hence position = 1.5t^2 i + (4 - 5t) j + (-1 -13t + t^2) k

b) speed v = 9t i - 5 j + (-13+2t) k

for slowest speed, we can ignore y speed value as it is not changing with time.

   speed v= sqrt(81t^2 + 5^2+4t^2 -52 t +169)

           = sqrt(85t^2-52 t+168)

dv/dt = 0, 170 t = 52

    t = 52/170 = 0.306 s

Minimum speed will be at t = 0.306 s

minimum speed = sqrt(81*0.306^2 + 5^2+4*0.306^2 -52*0.306 +169)

   = 13.64

speed at t = 1, v = sqrt(81*1^2 + 5^2+4*1^2 -52*1 +169)

                          =15.07

speed at t=4,   v = sqrt(81*4^2 + 5^2+4*4^2 -52*4 +169) = 36.69