Question

The position of a particle is given by r = 3.0t i -2.0t2 j + 4.0 k m.  where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the v and a of the particle? 

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

Answer 1

v(t)=(3.0 i-4.0t j); 𝑎⃗ =-4.0 j

a) The position of the particle is given by: 
r = 3.0t i -2.0t2 j + 4.0 k m
Velocity 𝑣⃗, of the particle is given as: 
v = \(\frac{dr}{dt}\) =\(\frac{ d}{dt}\) ( 3.0t i -2.0t2 j + 4.0 k m)
∴ v = 3.0 i - 4.0 t j
Acceleration 𝑎⃗, of the particle is given as:  
𝑎⃗ = \(\frac{dv}{dt} = \frac{d}{dt}\) ( 3.0 i - 4.0 t j)
∴ 𝑎⃗ -4.0j
8.54 m/s, 69.45° below the x-axis 

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b) We have velocity vector , v =  3.0 i - 4.0 t j
At t = 2.0 s;
v= 3.0 i - 8.0 t j

The magnitude of velocity is given by: 
\(|v| = \sqrt{ 3^2 (-8)^2 }\)= \(\sqrt{73}\) = 8.54 m/s

Direction, θ = tan-1 \((\frac{v_y}{ v_x} ) \)

= tan^-1 \((\frac{-8}{3})\)  = tan^-1 (2.667) 
= -69.25

The negative sign indicates that the direction of velocity is below the x-axis.