Re: Time taken to travel X miles is t seconds
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25 Oct 2025, 00:45
Given:
- Time to travel $x$ miles is $t$ seconds.
Quantity A:
- Time taken to travel 900 miles.
Quantity B:
- $\(15^{\frac{t}{x}}\)$.
Step 1: Express Quantity A in terms of $t$ and $x$
If traveling $x$ miles takes $t$ seconds, the time taken to travel 1 mile is $\(\frac{t}{x}\)$ seconds.
Therefore, the time taken to travel 900 miles is:
$$
\(900 \times \frac{t}{x}=\frac{900 t}{x}\)
$$
Thus,
$$
\(\text { Quantity } \mathrm{A}=\frac{900 t}{x}\)
$$
Step 2: Consider Quantity B
Quantity $\(\mathrm{B}=15^{\frac{t}{x}}\)$
Step 3: Compare Quantity A and Quantity B
- Quantity A is linear in $\(\frac{t}{x}\)$ (i.e., $\(900 \times \frac{t}{x}\)$ ),
- Quantity $B$ is exponential in $\(\frac{t}{x}\)$ (i.e., $\(15^{\frac{t}{x}}\)$ ).
Since we do not know the values of $t$ and $x$, the comparison depends on $\(\frac{t}{x}\)$.
- If $\(\frac{t}{x}\)$ is small (close to O ), Quantity A will be close to zero, and Quantity B will be close to $\(15^0=1\)$.
- If $\(\frac{t}{x}=1\)$, Quantity A is 900 and Quantity B is $\(15^1=15\)$, so Quantity A $\(>\)$ Quantity B.
- If $\(\frac{t}{x}\)$ is very large, Quantity A grows linearly, but Quantity B grows exponentially, so Quantity B will eventually be much larger.
Thus, without specific values for $t$ and $x$, no definite comparison can be made.
Conclusion:
The relationship between Quantity A and Quantity B depends on the value of $\(\frac{t}{x}\)$. No definite inequality can be established without further information.