Question

A 747 airliner reaches its takeoff speed of 173 mi/h in 35.2 s. what is the magnitude of its average acceleration

Answer 1

7.21 ft/s^2
       
Since you're looking for average acceleration, you can simply divide the change in velocity by the time. To make the calculation more reasonable, first convert the speed of 173 mi/h into ft/sec by multiplying by 5280 to convert from mi/h to ft/h and then dividing by 3600 to convert from ft/h to ft/s. 173 * 5280 / 3600 = 253.7333 ft/s Now divide the change in velocity by the time in seconds. 253.7333 ft/s / 35.2 s = 7.208333 ft/s^2 Rounding result to 3 significant figures gives 7.21 ft/s^2