Answer:
x_total = 600 m
Explanation:
This is an exercise and kinematics, let's find the time it takes to reach the velocity 20 m / s
v = v₀ + a t
as part of rest v₀ = 0
t = v / a
t = 20/2
t = 10 s
let's find the distance traveled in this time
x₁ = vo t + ½ a t2
x₁ = 0 + ½ 2 10²
x₁ = 100 m
The remaining time is
t₂ = 35 - t
t₂ = 35 - 10
t₂ = 25 s
as in this range it has a constant speed
v = x₂ / t₂
x₂ = v t₂
x₂ = 20 25
x₂ = 500 m
the total distance traveled is
x_total = x₁ + x₂
x_total = 100 + 500
x_total = 600 m
F=ma
Tension - weight = mass x acceleration
T - 5(9.81) = 5 x 1
T = 5 + 5(9.81)
T = 54.05 N
T ≈ 54 N
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
Water is, indeed, an exception. Normally when temperature drops, material shrinks. Water doesn't, water expands instead.
