Answer:
t₁ = 3 s
Explanation:
In this exercise, the vertical displacement equation is not given
y = 240 t + 16 t²
Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration
Let's replace
864 = 240 t + 16 t²
Let's solve the second degree equation
16 t² + 240 t - 864 = 0
Let's divide by 16
t² + 15 t - 54 = 0
The solution of this equation is
t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1
t = [-15 ±√(225 +216)] / 2
t = [-15 + - 21] / 2
We have two solutions.
t₁ = [-15 +21] / 2
t₁ = 3 s
t₂ = -18 s
Since time cannot have negative values, the correct t₁ = 3s
The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a
x₂ = 0.5*a*t² = 0.5*v°²/a
The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7
You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀.
Answer:
The correct answer is:
(A) to the left
(B) at speed -0.8725 m/s
Explanation:
The given values are:
Plate 1:
Mass,
m₁ = 201 g
Velocity,
v₁ = +1.79 m/s
Plate 2:
Mass,
m₁ = 335 g
Velocity,
v₁ = -2.47 m/s
According to the conservation of momentum, we get
⇒ 
then,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
⇒ 
(to the left)
