The position-time equation for a certain train is 2.9m + (8.8m/s)t + (2.4m/s2)+2 What is it’s acceleration?

Question

The position-time equation for a certain train is
2.9m + (8.8m/s)t + (2.4m/s2)+2

What is it’s acceleration?

in progress 0
Philomena 4 years 2021-07-20T19:32:53+00:00 1 Answers 17 views 0

Answers ( )

  1. RobertKer
    0
    2021-07-20T19:34:09+00:00
    Reply

    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.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )