The velocity graph of a particle moving along the x-axis is shown. The particle has zero velocity at t=0.00s and reaches a maximum velocity, vmax, after a total elapsed time, t total. If the initial position of the particle is x0 =9.35m, the maximum velocity of the particle is vmax=29.3m/s, and the total elapsed time is t total=29.9s, what is the particle's position at t=19.9s?
the graph is not shown.
the graph is a straight line
If the velocity versus time is straight, then its slope, the acceleration, is constant
a = change in v/change in t
= (29.3 - 0)/29.9
calculate a
v = Vi + a t
but Vi = 0
so at t = 19.9
v = 0 + a (19.9)
the graph is not shown.
To find the position of the particle at a specific time, we can use the velocity-time graph to determine the displacement of the particle.
Given that the particle starts at t=0.00s with a velocity of 0 m/s, we know that its initial position is x0 = 9.35m.
From the given information, we know that the maximum velocity of the particle is vmax = 29.3m/s, and the total elapsed time is ttotal = 29.9s.
To determine the position at t = 19.9s, we need to consider the behavior of the particle's velocity graph and integrate it to find the displacement.
Since the velocity-time graph is not provided, we can make some assumptions based on the given information:
1. The particle initially has zero velocity at t = 0.00s, meaning that its velocity starts from rest and increases.
2. The particle reaches its maximum velocity, vmax, after a total elapsed time, ttotal. Thus, the graph could be a linearly increasing function.
3. The velocity remains constant at vmax after reaching the maximum value.
Given these assumptions, we can break the problem into two parts:
1. Find the displacement from t = 0.00s to t = ttotal.
2. Find the additional displacement from t = ttotal to t = 19.9s.
1. The displacement from t = 0.00s to t = ttotal is the area under the velocity graph. Since the graph is a triangle with base ttotal and height vmax, we can use the formula for the area of a triangle:
Displacement1 = (1/2) * base * height = (1/2) * ttotal * vmax
2. The additional displacement from t = ttotal to t = 19.9s is the product of the velocity vmax and the time difference:
Displacement2 = vmax * (19.9s - ttotal)
To find the total displacement, we add Displacement1 and Displacement2:
Total Displacement = Displacement1 + Displacement2
Total Displacement = (1/2) * ttotal * vmax + vmax * (19.9s - ttotal)
Now, we can substitute the given values:
Total Displacement = (1/2) * 29.9s * 29.3m/s + 29.3m/s * (19.9s - 29.9s)
Total Displacement = 442.1m + 29.3m/s * (-10s)
Total Displacement = 442.1m - 293m
Total Displacement = 149.1m
Therefore, the particle's position at t = 19.9s is x0 + Total Displacement:
Position at t = 19.9s = x0 + Total Displacement
Position at t = 19.9s = 9.35m + 149.1m
Position at t = 19.9s = 158.45m
Thus, the particle's position at t = 19.9s is 158.45m.