Question

table
here is a table of rich’s and julio’s distance for the first few seconds.
what is your current prediction for the distance where rich will catch julio?
fill in some more rows if that helps to illustrate your thinking.
submit and explain

Explanation:

Step1: Find Julio's speed

Julio's distance at \( t = 1 \) is \( 4.8 \) yards, at \( t = 2 \) is \( 9.6 \) yards, at \( t = 3 \) is \( 14.4 \) yards. The speed \( v_{Julio} \) is constant (since \( \frac{9.6 - 4.8}{2 - 1}=\frac{14.4 - 9.6}{3 - 2}=4.8 \) yards per second). So Julio's distance formula is \( d_{Julio}(t)=4.8t \).

Step2: Find Rich's speed

Rich's distance at \( t = 1 \) is \( 2.07 \) yards, \( t = 2 \) is \( 8.27 \) yards, \( t = 3 \) is \( 14.47 \) yards. The difference between consecutive distances: \( 8.27 - 2.07 = 6.2 \), \( 14.47 - 8.27 = 6.2 \). Wait, at \( t = 0 \) and \( t = 1 \), the difference is \( 2.07 - 0 = 2.07 \), then \( 8.27 - 2.07 = 6.2 \), \( 14.47 - 8.27 = 6.2 \), \( 14.47 + 6.2 = 20.67 \) at \( t = 4 \)? Wait, no, maybe Rich's distance has an initial part and then constant speed. Wait, let's check the pattern again. Wait, Julio's distance at \( t = 1 \): \( 4.8 \), \( t = 2 \): \( 9.6 = 4.8\times2 \), \( t = 3 \): \( 14.4 = 4.8\times3 \), \( t = 4 \): \( 19.2 = 4.8\times4 \). Rich's distance: \( t = 0 \): \( 0 \), \( t = 1 \): \( 2.07 \), \( t = 2 \): \( 8.27 \) ( \( 2.07 + 6.2 \) ), \( t = 3 \): \( 14.47 \) ( \( 8.27 + 6.2 \) ), so from \( t = 1 \) onwards, Rich's speed is \( 6.2 \) yards per second? Wait, no, \( t = 1 \) to \( t = 2 \): \( 8.27 - 2.07 = 6.2 \), \( t = 2 \) to \( t = 3 \): \( 14.47 - 8.27 = 6.2 \), so Rich's distance formula: for \( t \geq 1 \), \( d_{Rich}(t)=d_{Rich}(1)+6.2(t - 1)=2.07 + 6.2(t - 1)=6.2t - 4.13 \). Let's check \( t = 2 \): \( 6.2\times2 - 4.13 = 12.4 - 4.13 = 8.27 \), correct. \( t = 3 \): \( 6.2\times3 - 4.13 = 18.6 - 4.13 = 14.47 \), correct. \( t = 4 \): \( 6.2\times4 - 4.13 = 24.8 - 4.13 = 20.67 \).

Step3: Set distances equal

We need to find \( t \) where \( d_{Julio}(t)=d_{Rich}(t) \). So \( 4.8t=6.2t - 4.13 \).

Step4: Solve for t

\( 6.2t - 4.8t = 4.13 \)

\( 1.4t = 4.13 \)

\( t=\frac{4.13}{1.4}\approx2.95 \) seconds? Wait, but let's check the table again. Wait, maybe I made a mistake in Rich's speed. Wait, at \( t = 0 \): 0, \( t = 1 \): 2.07, \( t = 2 \): 8.27, \( t = 3 \): 14.47, \( t = 4 \): 14.47 + 6.2 = 20.67. Julio at \( t = 3 \): 14.4, \( t = 4 \): 19.2. So at \( t = 3 \), Rich is at 14.47, Julio at 14.4. Close. At \( t = 4 \), Rich at 20.67, Julio at 19.2. Wait, maybe the initial difference: Rich at \( t = 1 \): 2.07, Julio at \( t = 1 \): 4.8. So Rich is behind. Then from \( t = 1 \), Rich's speed is \( (8.27 - 2.07)/(2 - 1)=6.2 \), Julio's speed is 4.8. So the relative speed is \( 6.2 - 4.8 = 1.4 \) yards per second. The distance Rich needs to cover to catch Julio is the distance between them at \( t = 1 \): Julio at \( t = 1 \) is 4.8, Rich at \( t = 1 \) is 2.07, so the gap is \( 4.8 - 2.07 = 2.73 \) yards? Wait, no, if Rich is faster than Julio from \( t = 1 \) onwards, then the time to catch up is the initial gap divided by relative speed. Wait, at \( t = 0 \), both at 0. At \( t = 1 \), Julio is at 4.8, Rich at 2.07. So Julio is ahead by \( 4.8 - 2.07 = 2.73 \) yards. Then Rich's speed is 6.2, Julio's is 4.8, so relative speed \( 6.2 - 4.8 = 1.4 \) yards per second. So time to catch up is \( 2.73 / 1.4 \approx 1.95 \) seconds after \( t = 1 \), so total time \( 1 + 1.95 = 2.95 \) seconds, which matches the earlier calculation. Let's check at \( t = 3 \): Rich is at 14.47, Julio at 14.4. So Rich has caught up around \( t = 3 \) seconds? Wait, the table shows at \( t = 3 \), Rich is 14.47, Julio is 14.4. So almost at \( t = 3 \) seconds. Let's check the exact time when \( d_{Rich}(t)=d_{Julio}(t) \).

\( d_{Rich}(t) \): for \( t \leq 1 \), maybe different, but from \( t = 1 \), \( d_{Rich}(t)=2.07 + 6.2(t - 1) \)

\( d_{Julio}(t)=4.8t \)

Set equal:

\( 2.07 + 6.2(t - 1)=4.8t \)

\( 2.07 + 6.2t - 6.2 = 4.8t \)

\( 6.2t - 4.13 = 4.8t \)

\( 1.4t = 4.13 \)

\( t = 4.13 / 1.4 = 2.95 \) seconds, approximately 3 seconds. Looking at the table, at \( t = 3 \), Rich is at 14.47, Julio at 14.4, so very close. At \( t = 2 \), Rich at 8.27, Julio at 9.6 (Julio is ahead). At \( t = 3 \), Rich at 14.47, Julio at 14.4 (Rich is ahead). So the catch-up happens between \( t = 2 \) and \( t = 3 \). Let's calculate at \( t = 2.95 \):

Julio: \( 4.8\times2.95 = 14.16 \)

Rich: \( 2.07 + 6.2\times(2.95 - 1)=2.07 + 6.2\times1.95 = 2.07 + 12.09 = 14.16 \). Yes, so at approximately \( t = 2.95 \) seconds, or about 3 seconds.

Answer:

Rich will catch Julio at approximately \( \boldsymbol{3} \) seconds (more precisely, around 2.95 seconds, but based on the table's pattern, at \( t = 3 \) seconds, Rich's distance (14.47 yards) is very close to Julio's (14.4 yards), so the catch - up occurs around 3 seconds. To fill the table, we can continue the patterns:

Time (seconds)	Rich (Yards)	Julio (Yards)
1	2.07	4.8
2	8.27	9.6
3	14.47	14.4
4	20.67	19.2
5	26.87	24.0

We can see that at \( t = 3 \) seconds, Rich's distance (14.47 yards) is just slightly more than Julio's (14.4 yards), so the catch - up happens at approximately 3 seconds.