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.