Question

a particle moves recently with a uniform acceleration of 2 metre per second square across a point a with a velocity of 5 m per second in the same direction as that of acceleration. find the velocity of the particle after one second ?

Answer 1

$a = (v - u) \div t$
'a' equals acceleration (2 m/s^2)
'u' equals initial velocity at point A (5 m/s)
't' equals time interval (1 sec after crossing point A)
'v' equals final velocity after 1 sec(to find)
therefore
v equals 7 m/s

Answer 2

a = (v - u) \div ta=(v−u)÷t 
'a' equals acceleration (2 m/s^2)
'u' equals initial velocity at point A (5 m/s)
't' equals time interval (1 sec after crossing point A)
'v' equals final velocity after 1 sec(to find)
therefore 
v equals 7 m/s