A car driving on a straight track accelerates at a rate of 3.2 m/s2 for 14 s. If the initial velocity of the car was 5.1 m/s, and its initia

Question

A car driving on a straight track accelerates at a rate of 3.2 m/s2 for 14 s. If the initial velocity of the car was 5.1 m/s, and its initial position was 0 m, what is its final position?

in progress 0
Thạch Thảo 1 month 2021-08-08T17:27:11+00:00 1 Answers 4 views 0

Answers ( )

    0
    2021-08-08T17:29:07+00:00

    Answer:

    The data we have is:

    The acceleration is 3.2 m/s^2 for 14 seconds

    Initial velocity = 5.1 m/s

    initial position = 0m

    Then:

    A(t) = 3.2m/s^2

    To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.

    V(t) = (3.2m/s^2)*t + 5.1 m/s

    To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)

    P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t

    Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.

    P(14s) =   (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m

    So the final position is 385 meters ahead the initial position.

Leave an answer

Browse

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