Answer:
time = 5.71 s
acceleration = -0.175 m/s²
v = 2.16 m/s
It is more than 2.1 m/s
Explanation:
given data
constant speed u = 2.6 m/s
distance s = 12 m
reduced constant speed v = 1.6 m/s
solution
we find how long it takes for the boat to coast the 12 meters
time = [tex]\frac{d}{0.5*(v+u)}[/tex]
time = [tex]\frac{12}{0.5*(1.6+2.6)}[/tex]
time = 5.71 s
we will apply equation of motion for During its coasting that is
v² - u² = 2as .........................1
here u is initial velocity i.e. 2.6 m/s and v is final velocity i.e 1.6 m/s and s is distance
1.6² - 2.6² = 2×a×12
a= -0.175 m/s²
and
When it coast 6 m then from equation 1
v² - u² = 2as
v² - 2.6² = 2×(-0.175)×12
v = 2.16 m/s
It is more than 2.1 m/s