A particle with initial velocity v₀ = (-2i - 4j) m/s undergoes a constant acceleration, a = 3 m/s² at 37° with positive direction of x - axis.
We have to find the velocity of particle at t = 5 sec.
initial velocity, v₀ = (-2i - 4j) m/s
constant acceleration, a = 3(cos37° i + sin37° j)
= 3(0.8 i + 0.6 j)
= 2.4 i + 1.8 j
let final velocity of the particle is v
time taken, t = 5 sec
now using formula,
[tex]\vec v=\vec u+\vec at[/tex]
here,
[tex]\vec u=v_0=(-2\hat i-4\hat j)\\\vec a=(2.4\hat i+1.8\hat j)[/tex]
[tex]\implies\vec v=(-2\hat i-4\hat j)+(2.4\hat i+1.8\hat j)\times5\\\\=10\hat i+5\hat j[/tex]
Therefore the velocity of particle at time t = 5sec is (10i + 5j)