course Phy 241 An automobile traveling a straight line is at point A at clock time t = 9 sec, where it is traveling at 10 m/s, to a certain point B. If the automobile accelerates uniformly at a rate of .9 m/s/s, then if it reaches point B at clock time t = 14 sec, what is its velocity at that point? What is its average velocity over the interval from A to B? If at point A the automobile is 44 meters from the starting point, how far is point B from the starting point? The final velocity is calculated by first finding `dt. This is calculated by 14-9, or 5 seconds. Now we can find the change in velocity during that time period by multiplying the 5 seconds by an acceleration of .9m/s^2. This gives 4.5 m/s. If the car starts at 10 m/s and it changes by 4.5 m/s, the final velocity must be 14.5 m/s. Average velocity is then calculated by adding the first velocity and the final velocity and dividing by 2, or (14.5 + 10)/2= 12.25 m/s. Before finding the distance from start to B, we can find out the distance from A to B. This quantity is represented by `ds. `ds is found by taking the average velocity and multiplying by `dt, or 12.25m/s * 5s=61.25m. If the distance from start to A is 44m, we can just add that to the distance from A to B to obtain the distance from start to B. This yields a distance of 105.25m.