Three cars (car F, car G, and car H) are moving with the same velocity, and slam on the brakes. The most massive car is car F, a
1 answer:
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Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t =
t =
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m
The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
- F₀ ≈ <u>0.377·M·g</u>
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Learn more about equilibrium of forces here: