A point moving on a straight horizontal

line with an initial velocity of 60 fps to the right
is given an acceleration of 12 fps2 to the left for
8 sec.
Determine (a) the total distance traveled during
the 8-sec interval; (b) the displacement during
the 8-sec interval.

To determine the total distance traveled during the 8-sec interval and the displacement during the same interval, we'll need to use kinematic equations.

(a) To find the total distance traveled:
1. First, we need to find the time at which the velocity becomes zero, as the point is decelerating due to the negative acceleration.
We can use the equation:
Final velocity (Vf) = Initial velocity (Vi) + (acceleration * time)
0 = 60 - 12 * time
Solving for time, we get:
time = 60 / 12 = 5 seconds

2. From this point, we can calculate the distance traveled during the acceleration phase.
Using the equation for distance with constant acceleration:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
Distance = (60 * 5) + (0.5 * -12 * 5^2)
Distance = 300 - 300 = 0 feet

3. As the velocity is now zero, the point starts accelerating in the opposite direction.
We can find the time at which the point reaches zero velocity again using the same equation as in step 1:
time = Final velocity (0) - Initial velocity (0) / acceleration (-12)
time = -0 / -12 = 0 seconds

4. Finally, we calculate the distance during this deceleration phase using the same equation as in step 2, but with the negative value of acceleration:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
Distance = (0 * 0) + (0.5 * -12 * 0^2)
Distance = 0 - 0 = 0 feet

Therefore, the total distance traveled during the 8-second interval is 0 + 0 = 0 feet.

(b) To find the displacement during the 8-sec interval:
Since displacement refers to the change in position irrespective of the path taken, we need to consider only the initial and final positions.

1. The initial position is given by the initial velocity * time:
Initial position = Initial velocity * time = 60 fps * 8 sec = 480 feet to the right.

2. The final position can be found using the equation for displacement with constant acceleration:
Displacement = (Initial velocity * time) + (0.5 * acceleration * time^2)
Displacement = (60 * 8) + (0.5 * -12 * 8^2)
Displacement = 480 - 384 = 96 feet to the right.

Therefore, the displacement during the 8-sec interval is 96 feet to the right.

At the beginning of the 8 s interval, the velocity is 60 ft/s.

At the end of the 8s interval, the velocity is 60 - 12*8 = -36 ft/s

The average velocity during the interval is (60-36)/2 = 12 ft/s
You can multiply that by 8 s for the total displacement.

(a) X = 60 t - 6 t^2
Maximum X value occurs @ dX/dt = 0, where t = 5 s. The value of X at that time is 300 - 150 = 150 ft. Then it goes backwards until t = 8, when
X(final) = 480 - 384 = 96 ft.

Total distance traveled, regardless of direction, is 150 + 96 = 246 ft

(b) Displacement from starting position = 96 feet. We already found that out part (a)