Question

A praticle with initial velocity v = -2i +4j m/s undergoes a constant acceleration a=3m/s^-2 at 37° with positive direction of x axis what is particle velocity after t=5sec​

Answer 1

Answer:

velocity after t=5sec​ V = 10i +13j

Explanation:

given

particle having initial velocity  v= -2i+4j m/s

having acceleration a = 3 m/s²

at angle 37 with x axis

divide a into x and y component

ax = 3×cos 37 = 3(4/5) = 12/5

ay = 3× sin 37 = 3(3/5) = 9/5

we have v= -2i + 4j

divide into x and y component

Vx= -2

Vy = 4

now take first equation of motion

V = U + at apply this into x and y direction

for X direction

Vx = Ux + ax t

Vx = (-2) + (12/5)5

Vx = (-2) + 12 = 10 m/sec in x direction

final velocity in x direction Vx = 10 m/s

for y direction

Vy = Uy + ay t

Vy = 4+ (9/5)5

Vy = 13 m/s in y direction

so we can write final velocity after 5 sec

V = Vx i + Vy j

V = 10i +13j