Write each force in component form:
<em>v </em>₁ : 50 N due east → (50 N) <em>i</em>
<em>v</em> ₂ : 80 N at N 45° E → (80 N) (cos(45°) <em>i</em> + sin(45°) <em>j</em> ) ≈ (56.5 N) (<em>i</em> + <em>j</em> )
The resultant force is the sum of these two vectors:
<em>r</em> = <em>v </em>₁ + <em>v</em> ₂ ≈ (106.5 N) <em>i</em> + (56.5 N) <em>j</em>
Its magnitude is
|| <em>r</em> || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N
and has direction <em>θ</em> such that
tan(<em>θ</em>) = (56.5 N) / (106.5 N) → <em>θ</em> ≈ 28.0°
i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive <em>x</em>-axis.)
The answer is D
Explanation:
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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Choice-C is a correct statement.
