course Phy 241 If an object increases velocity at a uniform rate from 7m/s to 21 m/s in 11s, what is its acceleration and how far does it travel? Sketch a velocity vs. time graph for an object whose initial velocity is 7m/s and whose velocity 11 sec later is 21 m/s. Explain what the slope of the graph means and why, and also what the area means and why. First we find the average acceleration by taking the change in velocity and divide by the time interval. (21m/s ?7 m/s)/11s=about 1.3 m/s^2. We can also find this by looking at the graph. The graph of this situation is linear and is increasing at a constant rate, because the velocity is increasing, but the acceleration is remaining constant. The acceleration can be found by the slope of the graph. This comes about because slope is defined by rise over run. The rise of the graph if velocity is on the y axis is the difference between the two velocities, and the run will by the time in seconds on the x axis. The rise over run is the same as the change in velocity over the change in time. To find the change in displacement, we must find the area under the velocity graph. The graph makes a trapezoid; therefore we can take the average height and multiply by the width to obtain `ds. The average height are the two velocities divided by two. The width is the time in seconds. Therefore we take [(21m/s + 7m/s)/2] *11 s=154m. We can check this with the formula, average velocity = change in s/change in t. This calculation from the graph is equivalent to the definition of average velocity.