Answer:
(a). The time is 26.67 sec.
(b). The distance traveled during this period is 1066.9 m.
Explanation:
Given that,
Speed = 80 m/s
Acceleration = 3 m/s
Initial velocity = 0
(a). We need to calculate the time
Using equation of motion
Put the value into the formula
The time is 26.67 sec.
(b). We need to calculate the distance traveled during this period
Using equation of motion
The distance traveled during this period is 1066.9 m.
Hence, This is the required solution.
Answer:
v = 19.33 m / s South
Explanation:
To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.
Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction
initial instant. Before the crash
p₀ = m 0 + M v₂ₓ
final instant. Right after the crash
p_f = (m + M) vₓ
p₀ = 0_pf
M v₂ₓ = (m + M) vₓ
In this case m is the mass of the car and M the mass of the SUV
vₓ = v₂ₓ (1)
in the Y axis (North - South direction)
initial instant
p₀ = m v_{1y} + M 0
final moment
p_f = (m + M) v_y
p₀ = p_f
m v_{1y} + M 0 = (m + M) v_y
v_y = (2)
With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.
W = ΔK
friction force work is
W = - fr d
the friction force is described by the equation
fr = μ N
Newton's second law
N-W = 0
N = W
we substitute
W = - μ (m + M) g d
as the car stops the final kinetic energy is zero and
the initial kinetic energy is
K₀ = ½ (m + M) v²
we substitute
- μ (m + M) g d = 0 - ½ (m + M) v²
μ g d = ½ v²
v² = 2 μ g d
the distance traveled can be found with the Pythagorean theorem
d =
d =
d = 8.40 m
let's calculate the speed
v² = 2 0.75 9.8 8.40
v = √123.48
v = 11.11 m / s
this velocity is in the direction of motion so we can use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (-5.48 / -6.37)
θ = 40.7º
Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.
θ'= 180 + 40.7
θ’= 220.7º
In the exercise they indicate the the sedan moves in the y-axis, therefore
sin θ'= v_y / v
v_y = v sin 220.7
v_y = 11.11 sin 220.7
v_y = -7.25 m / s
the negative sign indicates that it is moving south
To find the speed we substitute in equation 2
v_y =
v_{1y} = v_ y
let's calculate
v_{1y} = -7.25
v_{1y} = - 19.33 m/s
therefore the speed of the sedan is v = 19.33 m / s with a direction towards the South