The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in feet and t is measured in seconds. find the acceleration of the particle at 4 seconds. 3 ft/sec2 one third ft/sec2 negative 1 over 27 ft/sec2 none of these

Given that,

s = √ (2t + 1)

Time, t = 4 s

Acceleration, a = ?

Since,

Acceleration = velocity / time

Velocity = distance / time

Acceleration = distance / time²

s/t² = √ (2t+1) / t²

putting t = 4 sec, we have

a = √ (2*4+1) / 4²

a = √ (5) / 16

a = 0.139 ft/s²

Therefore, acceleration of the given particle will be 0.139 feet / second².