Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim
1 answer:
The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
Dividing we get,
Adding, we have,
Thus, the total time taken by Joshua is
To determine the average speed we use the formula,
where and
Hence, substituting the values we have,
Dividing, we get,
Thus, the average speed of Joshua during that time is 2500 m/h.
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<h2>Hello!</h2>
The graph is attached.
To graph a parabola we need to know the following:
- If the parabola is open upwards or downwards
- They axis intercepts (if they exist)
- The vertex position (point)
We are given the function:
Where,
For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.
Finding the axis interception points:
Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:
Using the quadratic equation:
So, at t=4.0615 the height of the softball will be 0.
Since we will work only with positive values of "x", since we are working with a function of time:
Let's start from "t" equals to 0 to "t" equals to 4.0615.
So, evaluating we have:
Finally, we can conclude that:
- The softball reach its maximum height at t equals to 2. (68 feet)
- The softball hits the ground at t equals to 4.0615 (0 feet)
- At t equals to 0, the height of the softball is equal to 4 feet.
See the attached image for the graphic.
Have a nice day!