Figure 2-41
gives the acceleration a versus time t for a particle moving along an x axis. The a-axis scale is set by as = 13.0 m/s2. At t = -2.0 s, the particle's velocity is 10.0 m/s. What is its velocity at t = 6.0 s?
what does the a-axis is "set" by as=13m/s^3 actually mean?
Is acceleration constant?
If acceleation is constant, then
v(6)=v(2)+a*t where t is six
Per the figure ( I was unable to post), the graph is a linear line, it doesn't mention that acceleration is constant.
To determine the velocity of the particle at t = 6.0 s, we need to analyze the given graph and use the information provided.
Figure 2-41 represents the acceleration (a) versus time (t) for the particle. Since the graph gives the acceleration, we need to integrate the acceleration over time to get the velocity.
Given that the x-axis scale is set by as = 13.0 m/s^2, it means that each unit on the x-axis represents 13.0 m/s^2.
First, we need to find the change in velocity (Δv) from t = -2.0 s to t = 6.0 s.
Δt = (6.0 s) - (-2.0 s) = 8.0 s
Since the acceleration (a) is constant throughout, the change in velocity is simply:
Δv = a * Δt
Given that as = 13.0 m/s^2, we can calculate the change in velocity:
Δv = (13.0 m/s^2) * (8.0 s) = 104.0 m/s
Next, we need to determine the initial velocity (v_0) at t = -2.0 s. Given that the velocity is 10.0 m/s at this time, v_0 = 10.0 m/s.
Finally, we can calculate the velocity (v) at t = 6.0 s using the formula:
v = v_0 + Δv
Substituting the values we determined:
v = 10.0 m/s + 104.0 m/s = 114.0 m/s
Therefore, the velocity of the particle at t = 6.0 s is 114.0 m/s.