Question

A hawk flies in a horizontal arc of radius 12.0 m at constant speed 4.00 m/s. (a) Find its centripetal acceleration. (b) It continues to fly along the same horizontal arc, but increases its speed at the rate of 1.20 m/s2. Find the acceleration (magnitude and direction) in this situation at the moment the hawk’s speed is 4.00 m/s.

Answer 1

Answer:

a) $a_c= 1.33 m/s^2$

b) a= 1.79 m/s²

   θ = 41.98⁰

Explanation:

arc radius  = 12 m

constant speed = 4.00 m/s

(a) centripetal acceleration

     $a_c=\frac{v^2}{R}$

     $a_c=\frac{4^2}{12}$

                  = 1.33 m/s²

(b) now we have given

        $a_t= \ 1.20 m/s^2$

        now,

         $a=\sqrt{a^2_c+ a^2_t}$

         $a=\sqrt{1.33^2+ 1.20^2}$

            a= 1.79 m/s²

 direction

$\theta = tan^{-1}(\frac{a_t}{a_r} )$

$\theta = tan^{-1}(\frac{1.2}{1.33} )$

     θ = 41.98⁰

Answer 2

The centripetal acceleration of the hawk is 1.33 m/s².

The resultant acceleration  of the hawk at the given moment is 1.79 m/s².

The direction resultant acceleration of the hawk is 48⁰.

The given parameters;

• radius of the arc, r = 12 m
• speed of the hawk, u = 4 m/s
• acceleration of the hawk, a = 1.2 m/s²

The centripetal acceleration of the hawk is calculated as follows;

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(4)^2}{12} \\\\a_c = 1.33 \ m/s^2$

The resultant acceleration is calculated as;

$a = \sqrt{a_c^2 + a_t} \\\\a = \sqrt{(1.33)^2 + (1.2)^2} \\\\a = 1.79 \ m/s^2$

The direction of the acceleration is calculated as follows;

$tan(\theta) = \frac{a_c}{a_t} \\\\\theta = tan^{-1} ( \frac{a_c}{a_t} )\\\\\theta = tan^{-1} ( \frac{1.33}{1.2} )\\\\\theta = 48^0$

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