To find the particle's position at t = 13.5 s, we need to analyze the given information and use the concepts of velocity and position.
Given data:
Initial position (xā) = 7.28 m
Maximum velocity (vāāā) = 33.4 m/s
Total elapsed time (t_total) = 20.3 s
To find the particle's position at t = 13.5 s, we need to break down the problem into two parts:
Part 1: Finding the displacement from t = 0.0 s to t = t_total
Part 2: Adding the displacement from Part 1 to the initial position (xā) to find the final position at t = t
Part 1: Finding the displacement from t = 0.0 s to t = t_total
The velocity graph provides information about the particle's velocity at different times. Since the particle starts at rest (zero velocity), the area under the velocity-time graph from t = 0.0 s to t_total gives the displacement of the particle during this time.
In this case, the velocity-time graph appears to be a straight line, indicating constant acceleration. The area under a velocity-time graph represents the displacement. Since it is a triangle, we can find the area using the formula for the area of a triangle:
displacement = (base * height) / 2
The base of the triangle is t_total, and the height can be calculated using vmax.
height = vmax
Using the given values, we can calculate the displacement:
displacement = (t_total * vmax) / 2
Part 2: Adding the displacement from Part 1 to the initial position (xā) to find the final position at t = t
To find the final position at t = t, we add the displacement from Part 1 to the initial position (xā).
final position = xā + displacement
Substituting the values we obtained:
final position = 7.28 m + displacement
Now, let's calculate the displacement and find the final position at t = 13.5 s:
displacement = (20.3 s * 33.4 m/s) / 2
final position = 7.28 m + displacement