A body moving in the positive x direction passes the origin at time t = 0. Between t = 0 and t = 1 second, thebody has a constant speed of 24 meters per second. At t = 1 second, the body is given a constant acceleration of6 meters per second squared in the negative x direction. The position x of the body at t = 11 seconds is(A) + 99m (B) + 36m (C) – 36 m (E) – 99 m

To solve this problem, we need to break it down into two parts: the first second of motion and the next 10 seconds of motion.

1. For the first second (t = 0 to t = 1), the body is moving at a constant speed of 24 meters per second. The distance covered in this time can be calculated using the formula for distance: distance = speed * time. So, the distance covered in the first second is 24 m/s * 1 s = 24 meters.

2. For the next 10 seconds (t = 1 to t = 11), the body is under a constant acceleration of -6 m/s² in the negative x direction. The displacement in this time can be calculated using the formula for motion under constant acceleration: displacement = initial velocity * time + 0.5 * acceleration * time². The initial velocity at t = 1 is the speed at that time, which is 24 m/s. So, the displacement in these 10 seconds is 24 m/s * 10 s + 0.5 * -6 m/s² * (10 s)² = 240 m - 300 m = -60 meters.

3. The total displacement of the body at t = 11 seconds is the sum of the displacements in the first second and the next 10 seconds. So, the position x of the body at t = 11 seconds is 24 m + (-60 m) = -36 meters.

Therefore, the correct answer is (C) -36 m.