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Jack is pushing a 30.0 kg cart at a constant acceleration of 2.1 m/s^2. He is combatting a force of 24N of gravity. If Jack let go of the cart when the velocity was 1.5 m/s, how far would the cart travel before stopping?

Respuesta :

Answer:

The car will travel 1,41 m before stop.

Explanation:

F = m * a

- 24 N (loss) = 30 Kg * a

a = -24/30 = -0,8 m/s^2

Using the Torricelli equation:

V^2 = Vo^2 = 2*a*S

final velocity = 0 (when the car stops)

initial velocity = 1,5 m/s (when the driver stops the acceleration)

0^2 m/s = (1,5 m/s)^2 + (2 * -0,8 m/s^2 * S)

0 = 2,25 - 1,6S

S = -2,25/-1,6

S = 1,41 m

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