Question

the bascule bridge shown is opening from its horizontal position to its fully vertical position. so far, the bridge has lifted 35° in 21 seconds. at this rate, how much longer will it take for the bridge to reach its vertical position?

s

Explanation:

Step1: Find the rate of lifting

The bridge lifts \( 35^\circ \) in \( 21 \) seconds. So the rate \( r \) is \( \frac{35^\circ}{21\ \text{s}}=\frac{5}{3}\ \text{degrees per second} \).

Step2: Calculate total time

To reach vertical, it needs to lift \( 90^\circ \) (from horizontal to vertical). Using time \( t=\frac{\text{total angle}}{\text{rate}} \), we have \( t = \frac{90^\circ}{\frac{5}{3}\ \text{degrees per second}} \).
Simplify: \( t=90\times\frac{3}{5}= 54 \) seconds. But wait, the bridge has already lifted for 21 seconds? Wait, no—wait, the question is "how much longer" so we need to find time to lift remaining angle. Wait, no, wait: the bridge is opening from horizontal to vertical. So total angle is \( 90^\circ \). It has lifted \( 35^\circ \) in 21s. Wait, no, maybe I misread. Wait, the problem says "how much longer will it take for the bridge to reach its vertical position?" So first, find the time to lift \( 90^\circ \) at the rate, then subtract the time already taken? Wait, no—wait, the rate is \( 35^\circ \) in 21s. So rate \( r=\frac{35}{21}=\frac{5}{3}\) degrees per second. The remaining angle is \( 90 - 35 = 55^\circ \)? Wait, no, wait—horizontal to vertical is \( 90^\circ \). If it has lifted \( 35^\circ \), then remaining is \( 90 - 35 = 55^\circ \)? Wait, no, maybe the bridge is lifting from horizontal (0°) to vertical (90°), so total angle to lift is 90°. It has lifted 35° in 21s. So time to lift 90° is \( \frac{90}{\frac{35}{21}}=\frac{90\times21}{35}= 54 \) seconds. Then the time already taken is 21s, so longer time is \( 54 - 21 = 33 \) seconds? Wait, no, wait—wait, maybe I made a mistake. Wait, let's re-express:

Rate: \( 35^\circ \) in 21s, so time per degree: \( \frac{21}{35}=\frac{3}{5}\) seconds per degree.

To lift 90°: total time \( 90\times\frac{3}{5}=54 \) seconds.

Time already taken: 21 seconds.

So longer time: \( 54 - 21 = 33 \) seconds. Wait, but let's check again.

Wait, the problem says "how much longer will it take for the bridge to reach its vertical position?" So the bridge is moving from horizontal (0°) to vertical (90°). It has moved 35° in 21s. So the remaining angle is \( 90 - 35 = 55^\circ \)? Wait, no, that can't be. Wait, maybe the bridge is lifting from horizontal to vertical, so the total angle to move is 90°. The rate is 35° per 21s. So time to move 90° is \( \frac{90}{35}\times21 = 54 \) s. So the time already spent is 21s, so the additional time is \( 54 - 21 = 33 \) s? Wait, but let's do it again.

Rate: \( \frac{35^\circ}{21\ \text{s}}=\frac{5}{3}\ \text{degrees per second} \).

Time to move 90°: \( t=\frac{90^\circ}{\frac{5}{3}\ \text{degrees per second}} = 90\times\frac{3}{5}= 54 \) s.

Time already taken: 21 s.

So longer time: \( 54 - 21 = 33 \) s.

Wait, but maybe the problem is that the bridge is lifting from horizontal to vertical, so the total angle is 90°, and we need to find the time to lift 90° at the rate, then that's the total time, but the question is "how much longer", so if it's already taken 21s, then 54 - 21 = 33. But let's check the calculation again.

\( \frac{35}{21}=\frac{5}{3} \) degrees per second.

Time for 90°: \( 90\div\frac{5}{3}=90\times\frac{3}{5}=54 \) s.

Longer time: 54 - 21 = 33 s.

Wait, but maybe I misread the problem. Let me re-read: "The bascule bridge shown is opening from its horizontal position to its fully vertical position. So far, the bridge has lifted \( 35^\circ \) in 21 seconds. At this rate, how much longer will it take for the bridge to reach its vertical position?"

Ah, so horizontal to vertical is 90°, so remaining angle is \( 90 - 35 = 55^\circ \)? Wait, no! Wait, 90 - 35 is 55? Wait, 90 - 35 = 55? Yes. Then time to lift 55° at rate \( \frac{35}{21}=\frac{5}{3}\) degrees per second. So time \( t=\frac{55}{\frac{5}{3}}=55\times\frac{3}{5}=33 \) s. Oh! That's a better way. So remaining angle is 90 - 35 = 55°, so time is \( \frac{55}{\frac{35}{21}}=\frac{55\times21}{35}=\frac{55\times3}{5}=33 \) s. Yes, that's correct. So that's the longer time.

Answer:

\( \boxed{33} \)