Consider an object moving in the positive direction which passes the point x = 0 at time t = 0. Between t = 0 and t = 6 seconds, the object has a constant velocity of +10 m/s. At time t = 6 seconds, the object is given a constant acceleration of 8 m/s2 in the negative direction. What is the position of the object at t = 16 seconds?

since s = vt + 1/2 at^2,

(10*6 + 0*6^2) + (10*10 + 4*10^2) = ____

560 m?

also shouldnt the acceleration be negative?

Yes, the 4 should have a - sign

To find the position of the object at t = 16 seconds, we can use the equations of motion.

First, let's find the position of the object from t = 0 to t = 6 seconds, when it has a constant velocity of +10 m/s. We can use the equation:

position = initial position + velocity * time

Since the initial position is x = 0, and the velocity is +10 m/s, we have:

position = 0 + 10 * t

For t = 6 seconds, the position is given by:

position = 0 + 10 * 6 = 60 meters

So, at t = 6 seconds, the object is at position 60 meters.

Now, let's find the position of the object from t = 6 seconds to t = 16 seconds, when it has a constant acceleration of -8 m/s^2. We can use the equation:

position = initial position + initial velocity * time + 0.5 * acceleration * time^2

Since we already know the initial position (60 meters), initial velocity (+10 m/s), and acceleration (-8 m/s^2), we can substitute these values into the equation:

position = 60 + 10 * t + 0.5 * (-8) * t^2

For t = 16 seconds, the position is given by:

position = 60 + 10 * 16 + 0.5 * (-8) * 16^2
position = 60 + 160 - 64 * 256
position = 60 + 160 - 16384
position = -16164 meters

So, at t = 16 seconds, the position of the object is -16164 meters.

Note: The negative sign indicates that the object is in the negative direction from the initial position at t=0.