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A driver loses control of a car, drives off an embankment, and lands in a canyon 6.0 m below. What was the car's speed just before touching the ground if it was traveling on the level surface at 12 m/s before the driver lost control?

Sagot :

leena

Hi there!

The horizontal component of the car's velocity (12 m/s) remains CONSTANT in the absence of air resistance.

Its vertical component of its velocity is being affected by gravity. We can use the kinematic equation:

[tex]v_f^2 = v_i^2 + 2ad[/tex]

vf = final velocity (? m/s)

vi = initial velocity (0 m/s)

a = acceleration due to gravity (9.8 m/s²)

d = displacement (6m)

Plug in the given values:

[tex]v_f^2 = 0 + 2(9.8)(6)\\\\v_f = 10.844 m/s \text{ downward}[/tex]

The speed is derived by using the pythagorean theorem as the above are simply the COMPONENTS of the total speed.

[tex]Speed = \sqrt{v_x^2 + v_y^2}\\\\s = \sqrt{12^2+10.844^2} = \boxed{16.174 m/s}[/tex]

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