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/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle?
a. 10 m/s
b. 16 m/s
c. 12 m/s
d. 14 m/s
e. 26 m/s

Answer 1

Answer:

e.26m/s

Explanation:

Vf=Vi+at      (1)

Vf=9j+(2i-4j)t

X= X₀+at

now, in the i direction

15=O+2t or t=7.5 when x position is 15

Lets put that into the (1) equation, solve for Vf.

 Vf=9j+(2i-4j)7.5

Vf= 15i - 21j

Speed= $\sqrt{xcomponent^2 + ycomponent^2}$

Vf= 25.8 m/s