I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.
Above
1 answer:
You might be interested in
Answer
given,
time = 10 s
ship's speed = 5 Km/h
F = m a
a is the acceleration and m is mass.
In the first case
F₁=m x a₁
where a₁ = difference in velocity / time
F₁ is constant acceleration is also a constant.
Δv₁ = 5 x 0.278
Δv₁ = 1.39 m/s
a₁ = 0.139 m/s²
F₂ =m x a₂
F₃ = F₂ + F₁
Δv₃ = 19 x 0.278
Δv₃ = 5.282 m/s
a₃=Δv₂ / t
a₃ = 0.5282 m²/s
m a₃=m a₁ + m a₂
a₃ = a₂ + a₁
0.5282 = a₂ + 0.139
a₂=0.3892 m²/s
F₂ = m x 0.3892...........(1)
F₁ = m x 0.139...............(2)
F₂/F₁
ratio =
ratio = 2.8
Answer:
d = 1.954 Km
Explanation:
given,
total distance, D = 2.5 Km
in stretch A to B =
speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t
in stretch B to C
time = 3.4 s
In stretch C to D
speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t
we know,
distance = speed x time
distance of BC
using equation of motion
v = u + a t
27.22 = 13.34 - a x 3.4
a = 4.08 m/s²
uniform deceleration is equal to 4.08 m/s²
distance traveled in BC
s = 68.94 m
3000 = 27.5 t + 68.94 + 13.33 t
40.83 t = 2931.06
t = 71.79 s
distance travel in AB
distance = s x t
d = 27.22 x 71.79
d = 1954 m
d = 1.954 Km
distance between A and B is equal to 1.954 Km.