A force (10i + 4j)N acts on a body of mass 2kg which is at rest. Find the velocity after 3 seconds.
\((\frac{5i}{3} + \frac{2j}{3})ms^{-1}\)
\((\frac{10i}{3} + \frac{4j}{3})ms^{-1}\)
\((5i + 2j)ms^{-1}\)
\((15i + 6j)ms^{-1}\)
Correct answer is D
Recall, \(F = mass \times acceleration \implies acceleration = \frac{force}{mass}\)
= \(\frac{10i + 4j}{2} = (5i + 2j) ms^{-2}\)
= \(v = u + at \implies v \text{at 3 seconds} = 0 + (5i + 2j \times 3)\)
= \((15i + 6j) ms^{-1}\)