For every _______ over 50 mph, your chances of being seriously injured, disfigured, or killed are doubled.
2 answers:
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Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :
Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt =
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :
For the position variation we use the vertical position equation :
Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is
For the second time interval :
t1 = 4 s → t2 = 9 s
Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :
Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then
The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
To get the velocity, we integrate over time:
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
To get the position equation we integrate again over time:
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
To get the position equation, we can integrate:
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
We can solve this for t.
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
We only care for the positive solution, which is:
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s