Well, it will surely be a piecewise function, so
d(t) =
3t for 0<=t<=1
3+2(t-1)/3 for 1<t<=4
5 for 4<t<=7
5+5(t-7)/3
Just look at the slope of each piece, and add the offset where it starts.
Ann went on a 10 hour bicycle trip. The table shows selected points on the graph.
t-0-1-4-7-10
d (t)- 0-3-5-5-10.
Write a function d(t) for her total distance traveled in miles in terms of the time in hours. Please help!
d(t) =
3t for 0<=t<=1
3+2(t-1)/3 for 1<t<=4
5 for 4<t<=7
5+5(t-7)/3
Just look at the slope of each piece, and add the offset where it starts.
Well, it looks like Ann's ride is like a roller coaster of emotions, or in this case, distances! Let's break it down:
For the first hour, Ann covered a grand total of 0 miles. She might as well have been riding on a stationary bike!
But then, in the second hour, she kicked things up a notch and traveled a distance of 3 miles. It's like she suddenly remembered she's not in a Tour de France race!
Then, in hours 3 to 6, Ann's progress seems to have hit a plateau. She maintained a constant speed and covered 5 miles each hour. Talk about finding your comfort zone!
But hold on, in the final hour, Ann went all out and reached a distance of 10 miles! She must have channeled her inner Olympic cyclist.
So, let's piece this all together into a function. We'll express the distance, d(t), where t is the time in hours:
d(t) = 0 for t ≤ 0,
d(t) = 3 for 0 < t ≤ 1,
d(t) = 5 for 1 < t ≤ 4,
d(t) = 5 for 4 < t ≤ 7,
d(t) = 10 for 7 < t ≤ 10.
So, there you have it. Ann's ride can be charted using this hilarious piecewise function. Enjoy the ups and downs (literally)!
For t = 0 to t = 1, the distance traveled is 3 - 0 = 3 miles.
For t = 1 to t = 4, the distance traveled is 5 - 3 = 2 miles.
For t = 4 to t = 7, the distance traveled is 5 - 5 = 0 miles.
For t = 7 to t = 10, the distance traveled is 10 - 5 = 5 miles.
Based on these intervals, we can write the piecewise function for the total distance traveled as follows:
d(t) = 3, for 0 ≤ t < 1
d(t) = 3 + 2(t - 1) = 2t + 1, for 1 ≤ t < 4
d(t) = 5, for 4 ≤ t < 7
d(t) = 5 + 5(t - 7) = 5t - 30, for 7 ≤ t ≤ 10
So, the function d(t) for her total distance traveled in terms of time in hours is:
d(t) = 3, for 0 ≤ t < 1
d(t) = 2t + 1, for 1 ≤ t < 4
d(t) = 5, for 4 ≤ t < 7
d(t) = 5t - 30, for 7 ≤ t ≤ 10
First, let's consider the time intervals between the given points. From t=0 to t=1, Ann traveled a distance of 3 miles (d=3-0=3). From t=1 to t=4, Ann traveled a distance of 2 miles (d=5-3=2). From t=4 to t=7, Ann traveled a distance of 0 miles (d=5-5=0). Finally, from t=7 to t=10, Ann traveled a distance of 5 miles (d=10-5=5).
We can observe that the distance traveled changes at specific time points and remains constant until the next time point.
Based on this observation, we can divide the 10-hour interval into multiple segments and define the distance traveled for each segment. Let's consider:
- For t between 0 and 1, the distance traveled is 3t. (d=3t)
- For t between 1 and 4, the distance traveled is 3 + 2(t-1). (d=3+2(t-1))
- For t between 4 and 7, the distance traveled is 5. (d=5)
- For t between 7 and 10, the distance traveled is 5 + 5(t-7). (d=5+5(t-7))
Now we can define the piecewise function d(t) as follows:
d(t) = {
3t, for 0 <= t < 1,
3 + 2(t-1), for 1 <= t < 4,
5, for 4 <= t < 7,
5 + 5(t-7), for 7 <= t <= 10
}
Therefore, the function for Ann's total distance traveled in terms of time is:
d(t) = {
3t, for 0 <= t < 1,
3 + 2(t-1), for 1 <= t < 4,
5, for 4 <= t < 7,
5 + 5(t-7), for 7 <= t <= 10
}
Note: Since the time interval is given as continuous, we assume that the function is continuous and the distance traveled smoothly changes between the given time points.