En un viaje ida y vuelta de 10 kilómetros, el desplazamiento y la distancia recorrida son respectivamente.
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Answer:
a) Stone v_{y} = - 7.25 ft / s , vₓ = 0.362 ft / s
b) tennis ball v_{y} = -3.16 ft / s , vₓ = 0.634 ft / s
c) golf ball v_{y} = - 1,536 ft / s, vx = 0.634 ft / s
2) golf ball
Explanation:
1) The average speed is defined with the displacement interval in the given time interval
v =( -x₀) / Δt
let's use this expression for each object
a) Stone
It tells us that it is released from y₀ = 10 ft and reaches the floor at
t = 0.788 s, but in the problem they tell us that the calculation is for t = 1.38 s
= (0-10) / 1.38
v_{y} = - 7.25 ft / s
in this interval a distance of x_{f} = 0.500 ft was moved away from the building (x₀ = 0 ft)
vₓ = (0.500- 0) / 1.38
vₓ = 0.362 ft / s
In my opinion it makes no sense to keep measuring the time after the stone has stopped.
b) tennis ball
It leaves the building at a height of y₀= 10ft and at the end of the period it is at a height of y_{f} = 5.63 ft, all this in a time of t = 0.788 + 0.591 = 1.38 s
the average vertical speed is
= (5.63 - 10) / 1.38
v_{y} = -3.16 ft / s
for horizontal velocity the ball leaves the building xo = 0 reaches the floor
x₁ = 0.500 foot and when bouncing it travels x₂ = 0.375 foot, therefore the distance traveled
x_{f} = x₁ + x₂
x_{f} = 0.500 + 0.375
x_{f} = 0.875 ft
we calculate
vₓ = (0.875 - 0) / 1.38
vₓ = 0.634 ft / s
c) The golf ball
the vertical displacement y₀ = 10 ft, and y_{f} = 7.88 ft
v_{y} = (7.88 - 10) / 1.38
v_{y} = - 1,536 ft / s
the horizontal displacement x₀ = 0 ft to the point xf = 0.875 ft
vₓ = (0.875 -0) / 1.38
vₓ = 0.634 ft / s
2) in this part we are asked for the instantaneous speed at the end of the time interval
a) the stone is stopped so its speed is zero
v_{y} = vₓ = 0
b) the tennis ball
It is at its maximum height so its vertical speed is zero
v_{y} = 0
horizontal speed does not change
vₓ = 0.634 ft / s
c) The golf ball
they do not indicate that it is still rising. Therefore its vertical speed is greater than zero
v_{y} > 0
horizontal speed is constant
vₓx = 0.634 ft / s
the total velocity of the object can be found with the Pythagorean theorem
v = √ (vₓ² + v_{y}²)
When reviewing the results, the golf ball is the one with the highest instantaneous speed at the end of the period
Here we have to calculate the acceleration of the wagon.
as per the question the mass of the wagon is given as 20 kg
the horizontal force that is applied on the body is 64 newton.
as the perfect unit of the fore is not mentioned it may be 64N or 64 dyne or 64 kgf.
as per Newton's second law of motion we know that the rate of change of momentum is the applied force.from the quantitative definition of Newton's second law we know that applied force is equal to the product of mass and acceleration.
hence
let F= 64 N,then
a=3.2 m/s^2
if F=64kgf=64×9.8 N,then
=31.36 m/s^2
if F =64 dyne ,the acceleration is found to be 0.0032 cm/s^2 [ans]