A cyclist whose mass including the bicycle is \(90 \mathrm{~kg}\) rides along a straight road for \(50 \mathrm{~s}\), as indicated in the figure. He then stops pedaling and slowly coasts to a stop. His average deceleration while stopping is nearest: (A) \(0.01 \mathrm{~m} / \mathrm{s}^2\) (B) \(0.05 \mathrm{~m} / \mathrm{s}^2\) (C) \(0.1 \mathrm{~m} / \mathrm{s}^2\) (D) \(0.5 \mathrm{~m} / \mathrm{s}^2\) (E) \(1 \mathrm{~m} / \mathrm{s}^2\)

We are given the mass of the cyclist including the bicycle (\(90 \mathrm{~kg}\)), the time duration of the ride (\(50 \mathrm{~s}\)), and the cyclist comes to a stop.

Since the cyclist comes to a complete stop, the final velocity will be \(0 \mathrm{~m}/\mathrm{s}\). We do not have enough information to directly calculate the initial velocity, so we need to make an assumption or estimation with the available data.

Let's assume that the cyclist has a constant speed during the \(50 \mathrm{~s}\) of riding, and that all the distance covered while pedaling is covered also in the same amount of time after the cyclist stops pedaling. This means that the initial velocity would be equal to the average speed during the entire ride.

Let's assume that the time it takes for the cyclist to come to a stop after pedaling is also \(50 \mathrm{~s}\), as there is no specific information given in the exercise.

Using the formula for average acceleration: \(a = \frac{Δv}{Δt}\), where \(a\) is average acceleration (deceleration in this case), \(Δv\) is the change in velocity, and \(Δt\) is the change in time. Since the initial and final velocities are equal, the change in velocity \(Δv = 0 - v_{initial}\). And given our assumption about the time to come to a stop, \(Δt = 50 \mathrm{~s}\).

Plug the values from Step 4 and Step 5 into the formula: \(a = \frac{Δv}{Δt} = \frac{0 - v_{initial}}{50 \mathrm{~s}}\). We still do not have the exact value for \(v_{initial}\), but we can analyze the most viable options.

Looking at the answer choices, we can see that some average deceleration values are too small or too large. For example, an average deceleration of \(1 \mathrm{~m}/\mathrm{s}^2\) will bring the cyclist to a stop very quickly, which is not the case here, so we can eliminate option (E). Now, let's explore the remaining choices and see how they affect the initial velocity: (A) \(0.01 \mathrm{~m}/\mathrm{s}^2\): \(v_{initial}= 0.01 \mathrm{~m}/\mathrm{s}^2 \times 50 \mathrm{~s} = 0.5 \mathrm{~m}/\mathrm{s}\) (B) \(0.05 \mathrm{~m}/\mathrm{s}^2\): \(v_{initial}= 0.05 \mathrm{~m}/\mathrm{s}^2 \times 50 \mathrm{~s} = 2.5 \mathrm{~m}/\mathrm{s}\) (C) \(0.1 \mathrm{~m}/\mathrm{s}^2\): \(v_{initial}= 0.1 \mathrm{~m}/\mathrm{s}^2 \times 50 \mathrm{~s} = 5 \mathrm{~m}/\mathrm{s}\) (D) \(0.5 \mathrm{~m}/\mathrm{s}^2\): \(v_{initial}= 0.5 \mathrm{~m}/\mathrm{s}^2 \times 50 \mathrm{~s} = 25 \mathrm{~m}/\mathrm{s}\) Given that the cyclist is riding casually and "slowly coasts to a stop," it is reasonable to assume that the cyclist's initial velocity is not as high as 25 \(m/s\), eliminating option (D). Based on this analysis, the most reasonable choice appears to be option (B), where the cyclist's initial velocity is 2.5 m/s, and the average deceleration is \(0.05 \mathrm{~m}/\mathrm{s}^2\).