Question

The position-time equation for a certain train is

Answer 1

Answer:

$a=4.8m/s^2$

Explanation:

Hello,

In this case, since the acceleration in terms of position is defined as its second derivative:

$a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)$

The purpose here is derive x(t) twice as follows:

$a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2$

Thus, the acceleration turns out 4.8 meters per squared seconds.

Best regards.