A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(37.00 \mathrm{~s}\) if its constant acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due east. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

To compute the accumulated velocity due to the acceleration, use the formula: \(V_{acc} = a \cdot t\) where \(V_{acc}\) is the accumulated velocity, \(a\) is the acceleration, and \(t\) is the time duration. Given that, \(a = 1.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(t = 37.00 \mathrm{~s}\), \(V_{acc} = 1.0 \mathrm{~m} / \mathrm{s}^{2} \cdot 37.00 \mathrm{~s} = 37.0 \mathrm{~m} / \mathrm{s}\)

We know that the initial velocity (\(V_{initial}\)) is \(20.0 \mathrm{~m} / \mathrm{s}\) west, and the accumulated velocity due to acceleration is \(37.0 \mathrm{~m} / \mathrm{s}\). Since both the initial velocity and the accumulated velocity due to acceleration are in opposite directions (west and east), we subtract the accumulated velocity from the initial velocity to find the final velocity: \(V_{final} = V_{initial} - V_{acc}\) \(V_{final} = 20.0 \mathrm{~m} / \mathrm{s} - 37.0 \mathrm{~m} / \mathrm{s}\) \(V_{final} = -17.0 \mathrm{~m} / \mathrm{s}\) Since the result is negative, it means the final velocity is in the opposite direction to the initial velocity, which is east. So, the final velocity of the car after \(37.00 \mathrm{~s}\) is \(17.0 \mathrm{~m} / \mathrm{s}\) east. The correct answer is option b) \(17.0 \mathrm{~m} / \mathrm{s}\) east.