Answer to Question #155452 in Physics for caroline chiok

Question #155452

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....


1
Expert's answer
2021-01-14T12:51:54-0500

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


v=v0atv = v_0 - at

where v0=30m/sv_0 = 30m/s is the initial speed, a=5m/s2a= 5m/s^2 is the acceleration, tt 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=10md = 10m travelled by the object is given as follows:


d=v0tat22d = v_0t - \dfrac{at^2}{2}

Solving the quadratic equation for tt, obtain:


t=v0+v022adat = \dfrac{v_0 + \sqrt{v_0^2 - 2ad}}{a}

Substituting into the first equation, obtain:


v=v0v0v022ad=v022adv=302251028.3m/sv = 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.


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