Question

A particle moves with an initial velocity 37 m/s and accelerates for 51 m/s^2 . Determine its velocity after 6 seconds

Answer 1

Answer:

The velocity after 6 seconds will be: v = 44.5 m/s

Explanation:

Initial velocity = u = 37 m/s

acceleration = a = 51 m/s²

Time = 6 seconds

Final velocity = v = ?

Using the equation of motion

v² = u²+2as

$v^2=\left(37\right)^2+2\left(51\right)\left(6\right)$

$v^2=1369+612$

$v=44.5$ m/s

Therefore, the velocity after 6 seconds will be: v = 44.5 m/s