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

---

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.