Answer to Question #155452 in Physics for caroline chiok

Question

An object initially traveling at 30. m/s south accelerates uniformly at 5.0 m/s/s north and is displaced 10 meters. The final velocity of the object is....

Answer 1

The velocity of an object under a constant acceleration is given as follows:

$v = v_0 - at$

where $v_0 = 30m/s$ is the initial speed, $a= 5m/s^2$ is the acceleration, $t$ is the time of motion. Sign minus here is due to the opposite directions of the initial velocity and acceleration (south and north respectively).

The distance $d = 10m$ travelled by the object is given as follows:

$d = v_0t - \dfrac{at^2}{2}$

Solving the quadratic equation for $t$, obtain:

$t = \dfrac{v_0 + \sqrt{v_0^2 - 2ad}}{a}$

Substituting into the first equation, obtain:

$v = v_0 - v_0-\sqrt{v_0^2 - 2ad} = -\sqrt{v_0^2 - 2ad}\\ v = -\sqrt{30^2 - 2\cdot 5\cdot 10} \approx -28.3m/s$

Answer. -28.3 m/s.