A particle starts from origin at t=0 with a velocity 5.0 î m/s and . . .

Question : A particle starts from origin at t=0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0i+2.0j) m/s². (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time?

Doubt by Saif

Solution :

u_x= 5 m/s
u_y= 0 m/s

a_x= 3 m/s²
a_y= 2 m/s²

x=84 m
t=?

y=?

a) We know,
S=ut+½at²

For Horizontal Motion

x=u_xt+½a_xt²
84=5t+½(3)t²
Multiply both sides by 2
[You can also take LCM and do the cross multiplication]

162 = 10t+3t²
3t²+10t-162=0

Factorising by splitting the Middle Term

3t²+(28-18)t-162=0
3t²+28t-18t-162=0
t(3t+28)-6(3t+28)=0
(t-6)(3t+28)=0
t-6=0 & 3t+28=0
t=6 & t=-28/3

But t≠-28/3
∴ t=6 sec

Now, 
For Vertical Motion 

y=u_yt+½a_yt²
y=(0)(6)+½(2)(6)²
y=0+36
y=36 m

Hence, the y-coordinate of the particle is 36 m when the x-coordinate is 84 m.  

b) We know,

v=u+at

For Horizontal Motion

V_x=u_x+a_xt
v_x=5+3(6)
v_x=5+18
v_x=23 m/s

For Vertical Motion

V_y=u_y+a_yt
V_y=0+2(6)
V_y=12 m/s

V = v_xi+v_yj
V=23i+12j


Speed of the particle
= |V|
= √[(23)²+(12)²]
= √[529+144]
= √[673]
= 25.94
= 26 m/s (approx)

Hence, the speed of the particle at t = 6s is 26 m/s (approx).