Question

A particle starts from the origin at t

t = 0 with an initial velocity of 4.9 m/s

m/s along the positive x

x axis.If the acceleration is (-2.9 i

^

i^ + 4.7 j

^

j^)m/s

2

m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x

x coordinate.

vx,vy =? rx,ry=?

Answer 1

$\vec a=-2.9\vec i+4.7\vec j$

If $t=0$ then $\vec v_0=4.9\vec i$ .

$\vec a=\frac{d\vec v}{dt}\to \vec v=\int\vec adt=(4.9-2.9t)\vec i+4.7t\vec j$

$\vec r=(4.9t-1.45t^2)\vec i+2.35t^2\vec j$

$r_x=r_{x \ max}$ when $t=1.69\ (s)$

(a) $v_x=4.9-2.9\cdot1.69=0$

$v_y=4.7\cdot1.69=7.9\ (m/s)$

(b) $r_{x\ max}=4.9\cdot 1.69-1.45\cdot1.69^2=4.14\ (m)$

$r_{y}=2.35\cdot1.69^2=6.7\ (m)$