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 = 14.0 m/s2. At t = -2.0 s, the particle's velocity is 11.0 m/s. What is its velocity at t = 6.0 s?
The hypothenuse of a right triangle measures to be 17.5 cm. If the three angles are 30 degrees, 60 degrees, and 90 degrees, what are the lengths of the other two sides?
To find the velocity of the particle at t = 6.0 s, we need to integrate the acceleration-time graph to obtain the velocity-time graph. By finding the area under the curve, we can determine the change in velocity.
First, let's determine the change in velocity from t = -2.0 s to t = 6.0 s.
The acceleration graph shows that the particle has a constant acceleration of 14.0 m/s². Since the acceleration is constant, this means that the velocity will change linearly over time.
Given that at t = -2.0 s, the velocity of the particle is 11.0 m/s, we can find the change in velocity by multiplying the acceleration by the time interval:
Change in velocity = acceleration * time interval
Change in velocity = 14.0 m/s² * (6.0 s - (-2.0 s))
Calculating the time interval:
6.0 s - (-2.0 s) = 8.0 s
Change in velocity = 14.0 m/s² * 8.0 s
Change in velocity = 112.0 m/s
The change in velocity from t = -2.0 s to t = 6.0 s is 112.0 m/s.
To find the velocity at t = 6.0 s, we add the change in velocity to the initial velocity:
Velocity at t = 6.0 s = Initial velocity + Change in velocity
Velocity at t = 6.0 s = 11.0 m/s + 112.0 m/s
Velocity at t = 6.0 s = 123.0 m/s
Therefore, the velocity of the particle at t = 6.0 s is 123.0 m/s.
To determine the velocity at t = 6.0 s, we need to integrate the given acceleration vs. time graph. Since the acceleration is constant between the two given points, we can use the kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
Initial velocity, u = 11.0 m/s
Acceleration, a = 14.0 m/s^2
Time, t = 6.0 s
Using the equation, we can calculate the final velocity:
v = u + at
v = 11.0 m/s + (14.0 m/s^2)(6.0 s)
v = 11.0 m/s + 84.0 m/s
v = 95.0 m/s
Therefore, the velocity of the particle at t = 6.0 s is 95.0 m/s.