A ship with an initial speed of 6.23 m/s approaches a dock that is 255 m away. If the ship accelerates uniformly and comes to rest in 82 s, what is
its acceleration?

a = 0.0761 [m/s^2]

Explanation:

This problem can be solved using kinematics equations, and by means of the following equation:

where:

Xo = initial location = 0

x = final location = 255 [m]

Vo = initial velocity = 6.23 [m/s]

t = time = 82 [s]

a = aceleration [m/s^2]

Since the ship slows down to rest, the last expression of the equation will be taken with a negative sign

255 = 0 + (6.23*82) - (0.5*a*82^2)

255 - (6.23*82) = - 3362*a

a = 0.0761 [m/s^2]

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//