Question

At t = 0, a particle leaves the origin with a velocity of 9 0 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (2.0i - 4.0j)m/s^2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle? 10m/S 16 m/s 12 m/s 14 m/s 26 m/s

Answer 1

acceleration
a=aₓ i + aᵧ j

where
aₓ = component of 'a' in X direction and
aᵧ = component of 'a' in Y direction

similarly for Velocity
u= uₓ i + uᵧ j

Now answer:

a= 2.0i - 4.0j
u= 0i + 9 j (at t=0 initial velocity)

aₓ= 2 m/s
aᵧ= -4 m/s

uₓ= 0 m/s
uᵧ= 9 m/s

x=15m
y=?

using s=ut+½at²

(in X direction)
x=u_xt+½a_xt²

=> 15= 0 + ½ (2t²)
=> t² = 15
=> t=3.87 sec

now final velocity of x direction

V²x = u_x² + 2a_xX

V²x = 0 + 2*2*15 = 60 m/s

in y direction

V_y = u_y+ a_y*t

V_y = 9 + (-4)*3.87 = -6.48

V =sqrt( V²_x + V²_y)

V = 10 m/s

option a is correct