The velocity v(t) of some particle is plotted
as a function of time on the graph below. The
scale on the horizontal axis is 1 s per division
and on the vertical axis 9 m/s per division.
Initially, the particle is at x0 = 52 m. What is the position x of the particle at
time t = 4 s?
Answer in units of m
test
87
To determine the position x of the particle at time t = 4s, we need to find the area under the velocity-time graph up to t = 4s and add it to the initial position x0.
Looking at the graph, we can divide it into rectangles and triangles. The area under the graph can be calculated by finding the sum of the areas of these shapes.
Let's calculate the areas of each shape:
1. Rectangle 1: The width of the rectangle is 1s, and the height is 9m/s. Therefore, the area of Rectangle 1 is 1s * 9m/s = 9m.
2. Triangle 1: The base of the triangle is 1s, and the height is -6m/s (since the velocity is negative). The area of Triangle 1 is 0.5 * 1s * (-6m/s) = -3m.
3. Triangle 2: The base of the triangle is 2s, and the height is -9m/s. The area of Triangle 2 is 0.5 * 2s * (-9m/s) = -9m.
4. Rectangle 2: The width of the rectangle is 1s, and the height is -3m/s. Therefore, the area of Rectangle 2 is 1s * (-3m/s) = -3m.
Now, let's find the total area:
Total area = Area of Rectangle 1 + Area of Triangle 1 + Area of Triangle 2 + Area of Rectangle 2
= 9m + (-3m) + (-9m) + (-3m)
= 9m - 3m - 9m - 3m
= -6m
Finally, we can calculate the position x of the particle at t = 4s as follows:
x = x0 + Total area
= 52m + (-6m)
= 46m
Therefore, the position x of the particle at t = 4s is 46m.
To determine the position of the particle at time t = 4s, we will need to find the area under the velocity vs. time graph from t = 0s to t = 4s.
Looking at the graph, we see that the velocity of the particle is increasing linearly from t = 0s to t = 1s. The velocity during this interval can be calculated by finding the slope of the line connecting the two points on the graph at t = 0s and t = 1s.
The slope can be calculated as (change in velocity) / (change in time) = (9 m/s - 0 m/s) / (1 s - 0 s) = 9 m/s.
Now, we can find the area of the triangle representing the velocity vs. time graph from t = 0s to t = 1s as: (base * height) / 2 = (1s * 9 m/s) / 2 = 4.5 m.
Next, we observe that the velocity of the particle remains constant at 9 m/s from t = 1s to t = 3s. Therefore, the area under this portion of the graph is simply: velocity * time = 9 m/s * (3s - 1s) = 18 m.
Lastly, from t = 3s to t = 4s, the velocity is decreasing linearly from 9 m/s to 0 m/s. Again, we calculate the slope of the line connecting these two points as: (0 m/s - 9 m/s) / (4 s - 3 s) = -9 m/s.
Finding the area under the triangle from t = 3s to t = 4s yields: (base * height) / 2 = (1s * 9 m/s) / 2 = 4.5 m.
To find the total displacement or position x of the particle, we sum up the three areas:
Displacement = 4.5 m + 18 m + 4.5 m = 27 m
Therefore, the position x of the particle at time t = 4s is 27 m.