Answer:
216.59 s.
Explanation:
Using,
F = ma................. Equation 1
Where F = force exerted by the tugboat, m = mass of the ocean liner, a = acceleration of the ocean liner
make a the subject of the equation
a = F/m.................. Equation 2
Given: F = 195 kN = 195000 N, m = 38000 Mg = 38000000 kg.
Substitute into equation 2
a = 195000/38000000
a = 5.13×10⁻³ m/s²
Also using,
Assuming the liner is decelerating
a = (u-v)/t............ Equation 3
Where v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (u-v)/a............. Equation 4
Given: u = 4 km/h = 4(1000/3600) = 1.111 m/s, v = 0 m/s, a = 0.00513 m/s²
Substitute into equation 4
t = (1.111-0)/0.00513
t = 216.59 s.
