To find the train's acceleration, we can use the equations of motion. Let's assume the initial velocity of the train is zero.
First, let's find the acceleration during the first second:
Using the equation: s = ut + (1/2)at^2, where s = displacement (10 m), u = initial velocity (0 m/s), t = time (1 s), and a = acceleration (to be determined).
Plugging in the values, we get:
10 = 0(1) + (1/2)a(1)^2 10 = 0 + (1/2)a 20 = a
So, the acceleration during the first second is 20 m/s^2.
Next, let's find the acceleration during the next two seconds:
Using the same equation: s = ut + (1/2)at^2, where s = displacement (32 m), u = initial velocity (0 m/s), t = time (2 s), and a = acceleration (to be determined).
Plugging in the values, we get:
32 = 0(2) + (1/2)a(2)^2 32 = 0 + 2a 16 = a
So, the acceleration during the next two seconds is 16 m/s^2.
Since the train has a uniform acceleration, we can take the average of the two accelerations to find the overall acceleration:
Average acceleration = (20 + 16) / 2 = 36 / 2 = 18 m/s^2.
Therefore, the train's acceleration is 18 m/s^2, not 4 as mentioned in the book.