The F-14 Tomcat lands on aircraft carriers at a speed of 145 mph. The carriers capture system stops the aircraft within 2 seconds. What is the F-14's acceleration rate?
in m/s^2?
Convert 145 mph to m/s and divide that answer by 2 seconds.
To find the acceleration rate, we will use the following formula:
acceleration (a) = (final velocity (vf) - initial velocity (vi)) / time (t)
First, we need to convert the speed of the F-14 Tomcat from mph to m/s:
145 mph = 145 * 0.44704 m/s
≈ 64.8518 m/s
Next, we can substitute the given values into the formula:
vf (final velocity) = 0 m/s (because the F-14 Tomcat comes to a stop)
vi (initial velocity) = 64.8518 m/s
t (time taken) = 2 seconds
Therefore:
acceleration (a) = (0 m/s - 64.8518 m/s) / 2 s
= -64.8518 m/s / 2 s
≈ -32.4259 m/s^2
The F-14 Tomcat's acceleration rate is approximately -32.4259 m/s^2. Note that the negative sign indicates that the plane is slowing down.