A particle starts from xi = 10 m at t0 = 0 and moves with the velocity graph shown in the graph below.

Sorry for no graph it's at 12 m/s when time is zero. Moves down consistently so that it is 8 at 1 4 at 2 0 at 3 and negative 2 at 4.

(a) What is the object's position at t = 2 s, 3 s, and 4 s?
m (at t = 2 s)
m (at t = 3 s)
m (at t = 4 s)

(b) Does this particle have a turning point? If so, at what time?
s

(a) Let's calculate the position of the object at different times using the given velocity graph:

At t = 2s:
The object's velocity is 4 m/s (negative direction). Starting from xi = 10 m, the object travels for 2 seconds with a velocity of -4 m/s. Using the equation x = xi + v*t, we can find the position:
x = 10 + (-4)*2 = 10 - 8 = 2 m

At t = 3s:
The object's velocity is 0 m/s, indicating that it momentarily stops. Using the same equation, we find the position at t = 3s:
x = 2 + 0*3 = 2 m

At t = 4s:
The object's velocity is -2 m/s. Using the equation, we find the position:
x = 2 + (-2)*4 = 2 - 8 = -6 m

So, the positions at t = 2s, t = 3s, and t = 4s are 2 m, 2 m, and -6 m, respectively.

(b) Yes, the particle has a turning point(s) when its velocity changes direction from negative to positive or vice versa. Looking at the velocity graph, we can see that the particle changes its direction at t = 3s, where the velocity goes from negative to zero. Hence, t = 3s is the turning point.

(a) To find the object's position at different times, we can integrate the velocity graph with respect to time. Since the velocity graph is given piecewise, we need to break it down into different intervals.

For t = 0 to t = 1:
The object's velocity is constant at 12 m/s, so the displacement during this interval is:

∆x1 = velocity * time = 12 m/s * 1 s = 12 m

For t = 1 to t = 2:
The object's velocity is decreasing linearly, starting at 12 m/s and reaching 4 m/s at t = 2. Since the velocity is changing linearly, we can use the average velocity during this interval to find the displacement:

∆x2 = average velocity * time = (12 m/s + 4 m/s) / 2 * 1 s = 8 m

For t = 2 to t = 3:
The object's velocity is constant at 4 m/s, so the displacement during this interval is:

∆x3 = velocity * time = 4 m/s * 1 s = 4 m

For t = 3 to t = 4:
The object's velocity is decreasing linearly, starting at 4 m/s and reaching -2 m/s at t = 4. We can again use the average velocity during this interval to find the displacement:

∆x4 = average velocity * time = (4 m/s + (-2 m/s)) / 2 * 1 s = 1 m

The object's position at t = 2 s is given by:
xi + ∆x1 + ∆x2 = 10 m + 12 m + 8 m = 30 m

The object's position at t = 3 s is given by:
xi + ∆x1 + ∆x2 + ∆x3 = 10 m + 12 m + 8 m + 4 m = 34 m

The object's position at t = 4 s is given by:
xi + ∆x1 + ∆x2 + ∆x3 + ∆x4 = 10 m + 12 m + 8 m + 4 m + 1 m = 35 m

Therefore, the object's positions at t = 2 s, 3 s, and 4 s are 30 m, 34 m, and 35 m, respectively.

(b) To determine if the particle has a turning point, we need to identify when the velocity changes from decreasing to increasing (or vice versa).

From the velocity graph, we can see that the velocity starts at 12 m/s and decreases until it reaches -2 m/s at t = 4 s. Therefore, there is a turning point at t = 4 s, where the particle changes direction.

Therefore, the particle has a turning point at t = 4 s.

To determine the position of the particle at different times, we need to integrate the velocity graph with respect to time. The integral of velocity gives us the displacement, which represents the change in position.

(a) To calculate the position at a particular time, we need to integrate the velocity function. Looking at the graph you described, the velocity is constant at 12 m/s for the initial time period. Let's call this time period T1.

In T1, we have:
Velocity = 12 m/s
Time = 1 s (from t = 0 s to t = 1 s)

Using the equation: Displacement = Velocity × Time, we can find the displacement during this time interval:
Displacement in T1 = 12 m/s × 1 s = 12 m

So at t = 1 s, the particle's position is 12 meters.

Next, let's consider the time period where the velocity linearly decreases from 12 m/s to 0 m/s. Let's call this time period T2.

In T2, the velocity decreases by 4 m/s per second. So, in 2 seconds (from t = 1 s to t = 3 s), the velocity decreases by 8 m/s.

Therefore, the velocity at t = 3 s is 12 m/s - 8 m/s = 4 m/s.

Using the same equation as before, we can find the displacement during this time interval:
Displacement in T2 = 4 m/s × 2 s = 8 m

So at t = 3 s, the particle's position is 12 m + 8 m = 20 meters.

Finally, we have the time period where the velocity is constant at 0 m/s. Let's call this time period T3.

In T3, the particle's velocity is 0 m/s, which means there is no change in position. Therefore, at t = 4 s, the particle's position is the same as at t = 3 s, which is 20 meters.

So, the positions of the particle at t = 2 s, 3 s, and 4 s are:
- At t = 2 s: 12 m
- At t = 3 s: 20 m
- At t = 4 s: 20 m

(b) To find the turning point of the particle, we need to identify when the velocity changes direction. Based on the information given, the velocity graph starts with a positive value (12 m/s) and consistently decreases until it becomes negative (-2 m/s).

Therefore, the particle changes direction at t = 4 s, where the velocity changes from positive to negative. This turning point represents the time when the particle changes direction.

(a) 24, 18, 32

(b) yes, and -4