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?

Answer 1

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.