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.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

80cm/s.

24cm/s.

53.3cm/s.