A train is traveling south at 25.6 m/s when the brakes are applied. It slows down with a constant acceleration to a speed of 6.90 m/s in a time of 8.38 s. What is the acceleration of the train during the 8.38-s interval?
a = ∆v/∆t
= (6.90-25.6)m/s / (8.38-0)s
= (-18.7 m/s)/8.38s
= -2.23 m/s^2
To find the acceleration of the train during the 8.38-second interval, you can use the equation for acceleration:
acceleration = (final velocity - initial velocity) / time
Given information:
Initial velocity (u) = 25.6 m/s (train's speed before the brakes are applied)
Final velocity (v) = 6.90 m/s (train's speed after 8.38 seconds)
Time (t) = 8.38 s
Plugging in the values into the equation, we get:
acceleration = (6.90 m/s - 25.6 m/s) / 8.38 s
Now, let's solve the equation to find the acceleration:
acceleration = (-18.7 m/s) / 8.38 s
acceleration = -2.23 m/s^2
Therefore, the acceleration of the train during the 8.38-second interval is -2.23 m/s^2. The negative sign indicates that the train is decelerating or slowing down.