Question

A particle P has a velocity (-3i+j)m/s at a time t=0,the particle moves with constant acceleration a=(2i+3j)m/s^2.find the speed of the particle after 3secs.

Answer 1

Using the formula

$v=v_0+a\cdot t$

We write the speed formulas for each component after 3 seconds

$v_i=v_{0i}+a_i\cdot t=-3+2\cdot t=-3+2\cdot 3=-3+6=3 m/s$

$v_j=v_{0j}+a_j\cdot t=1+3\cdot t=1+3\cdot 3=1+9=10 m/s$

Then the velocity modulus is

$v=\sqrt{v_i^2+v_j^2}=\sqrt{3^2+10^2}=\sqrt{9+100}=\sqrt{109}=10.44m/s$