The figure gives the acceleration a versus time t for a particle moving along an x axis. The a-axis scale is set by as = 15.0 m/s^2. At t = -2.0 s, the particle's velocity is 7.00 m/s. What is its velocity at t = 6.0 s?
To find the velocity of the particle at t = 6.0 s, we need to integrate the acceleration versus time graph given in the figure.
First, observe that the a-axis scale is set by as = 15.0 m/s^2. This means that the acceleration is scaled such that each grid line represents an acceleration of 15.0 m/s^2. From the graph, we can see that at t = -2.0 s, the acceleration is 3 grid lines below the zero line. Therefore, the acceleration at t = -2.0 s is -3 * 15.0 m/s^2 = -45.0 m/s^2.
Next, we need to compute the change in velocity of the particle from t = -2.0 s to t = 6.0 s. To do this, we can use the definition of acceleration:
a = Δv / Δt,
where a is the acceleration, Δv is the change in velocity, and Δt is the change in time.
Rearranging the equation, we have:
Δv = a * Δt.
Substituting the given values, we have:
Δv = (-45.0 m/s^2) * (6.0 s - (-2.0 s)).
Simplifying, we get:
Δv = (-45.0 m/s^2) * (8.0 s).
Δv = -360.0 m/s.
Therefore, the change in velocity is -360.0 m/s.
To find the velocity at t = 6.0 s, we need to add this change in velocity to the initial velocity at t = -2.0 s. From the problem, we know that at t = -2.0 s, the particle's velocity is 7.00 m/s. Adding the change in velocity:
velocity at t = 6.0 s = 7.00 m/s + (-360.0 m/s) = -353.0 m/s.
Therefore, the velocity of the particle at t = 6.0 s is -353.0 m/s.