An automobile whose speed is increasing at a rate of 0.600 m/s2 travels along a circular road of radius 30.0 m.

(a) When the instantaneous speed of the automobile is 5.00 m/s, find the tangential acceleration component.
Magnitude
Direction

in a direction opposite to the automobile's motionin the direction of the automobile's motion

(b) What is the centripetal acceleration component?
Magnitude
Direction

toward the center of curvatureaway from the center of curvature

(c) Determine the magnitude and direction of the total acceleration.
Magnitude

Direction

Question

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moya1316 moya1316 · Answer 1 · 2019-11-16T18:35:22+01:00

Answer:

a) [tex]a_{t}[/tex] = 0.600 m /s² , The direction is tangential to the curve at all times.

b) [tex]a_{c}[/tex] = 0.833 m / s² , to center radius

c) a = 1,027 m / s² , θ = 54.2º

Explanation:

In this exercise we have two accelerations a tangential acceleration and a centripetal acceleration.

a = [tex]a_{t}[/tex] + [tex]a_{c}[/tex]

Where centripetal acceleration is given by

[tex]a_{c}[/tex] = v² / r

a) as they give us the acceleration for the change of speed this is the tangential acceleration

[tex]a_{t}[/tex] = 0.600 m / s2

The direction is tangential to the curve at all times.

b) centripetal acceleration is

[tex]a_{c}[/tex] = v² / r

[tex]a_{c}[/tex] = 5²/30

[tex]a_{c}[/tex] = 0.833 m / s²

c) total acceleration

Let's use Pythagoras' theorem for total acceleration, since the two are perpendicular

a² = [tex]a_{t}[/tex]² + [tex]a_{c}[/tex] ²

a = √ ([tex]a_{c}[/tex] ² + (v² / r)²

a = √ (0.600² + (0.833)²

a = √ (0.36 + 0.694)

a = 1,027 m / s²

For the angle we use trigonometry

tan θ = [tex]a_{c}[/tex] /[tex]a_{t}[/tex]

θ = tan⁻¹ ([tex]a_{c}[/tex] /[tex]a_{t}[/tex] )

θ= tan⁻¹ (0.833 / 0.600)

θ = 54.2º