Suppose that an accelerating car goes from 0 mph to 68.2 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 feet per sec = 22/15 ft/s.) Find the average acceleration of the car over each of the first two seconds.
t-0,1,3,4,5
v(t)-0.00,34.09,59.09,77.27,90.91,100.00
A.)average acceleration over the first second =
B.)average acceleration over the second second=
A) The change in velocity over the first second is v(1) - v(0) = 34.09 - 0.00 = 34.09 ft/s. The time interval is 1 second. Therefore, the average acceleration over the first second is:
a = (v(1) - v(0)) / (1 - 0) = 34.09 ft/s^2
B) The change in velocity over the second second is v(2) - v(1) = 59.09 - 34.09 = 25.00 ft/s. The time interval is 1 second. Therefore, the average acceleration over the second second is:
a = (v(2) - v(1)) / (2 - 1) = 25.00 ft/s^2