Question

what is the distance traveled by the car in the 20th second?
a) 100 m
b) 20 m
c) 10 m
d) 40 m

Explanation:

To solve the problem of finding the distance traveled by the car in the 20th second, we need to assume the motion of the car. Usually, for such problems, if the car is moving with a constant acceleration (or sometimes constant velocity, but the "nth second" distance often involves uniformly accelerated motion), we use the formula for the distance traveled in the nth second:

The formula for the distance traveled in the \( n^{\text{th}} \) second is given by:

\[ s_n = u + \frac{a}{2}(2n - 1) \]

where \( u \) is the initial velocity, \( a \) is the acceleration, and \( n \) is the time in seconds.

However, since the problem is not providing the initial velocity (\( u \)) and acceleration (\( a \)) directly, we might assume a common case (like starting from rest, \( u = 0 \)) and a constant acceleration (for example, if the car is moving with \( a = 2 \, \text{m/s}^2 \), but this is an assumption as the problem is incomplete). Wait, maybe the problem is related to a car moving with a constant velocity? No, the "nth second" distance is more related to uniformly accelerated motion.

Wait, maybe there is a missing part of the problem. But since the options are given, let's think differently. If we assume the car is moving with a constant velocity of \( v = 10 \, \text{m/s} \), then the distance traveled in any second would be \( v \times 1 \, \text{s} = 10 \, \text{m} \). But that might not be the case.

Alternatively, if the car is moving with an acceleration such that the distance in the nth second is \( 10 \, \text{m} \) when \( n = 20 \). Wait, maybe the problem is about a car moving with \( u = 0 \) and \( a = 1 \, \text{m/s}^2 \). Then the distance in the nth second is \( s_n = 0 + \frac{1}{2}(2n - 1) \). For \( n = 20 \), \( s_{20} = \frac{1}{2}(40 - 1) = \frac{39}{2} = 19.5 \, \text{m} \), which is close to 20 m. But this is an assumption.

Alternatively, if the car is moving with a constant velocity of \( 20 \, \text{m/s} \), the distance in 1 second is \( 20 \, \text{m} \), but that would be option B. But this is all guesswork without the full problem.

Wait, maybe the original problem had the car starting from rest and accelerating at \( 1 \, \text{m/s}^2 \), but no. Alternatively, maybe the car is moving with a velocity of \( 10 \, \text{m/s} \), so distance in 1 second is \( 10 \, \text{m} \), which is option C.

But since the problem is incomplete, but the options are given, and assuming a common problem where the distance in the nth second is calculated, and if we consider the formula \( s_n = u + \frac{a}{2}(2n - 1) \), and if we assume \( u = 0 \) and \( a = 1 \, \text{m/s}^2 \), then for \( n = 20 \), \( s_{20} = \frac{1}{2}(39) = 19.5 \approx 20 \, \text{m} \), so option B. But this is an assumption.

Alternatively, if the car is moving with a constant velocity of \( 10 \, \text{m/s} \), then distance in 1 second is \( 10 \, \text{m} \), option C.

But without the full problem, it's hard to say. However, looking at the options, the most probable answer (assuming a common problem) is:

Assuming the car is in uniformly accelerated motion (e.g., starting from rest with appropriate acceleration) or constant velocity, the distance in the 20th second is calculated. Using the formula for distance in the \( n^{\text{th}} \) second (\( s_n = u + \frac{a}{2}(2n - 1) \)) or constant velocity logic, the answer aligns with option B (20 m) as a common result for such problems.

Answer:

B) 20 m