A cyclist is riding with a speed of 36 km /h on a circular road of radius 100m. He applied brakes and reduces his speed at the constant rate of 0.75m/s every second. The angle between net acceleration and velocity at that instant is

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A cyclist is riding with a speed of 36 km /h on a circular road of radius 100m. He applied brakes and reduces his speed at the constant rate of 0.75m/s every second. The angle between net acceleration and velocity at that instant is

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127°

143°

37°

30°

answer is D.

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Detailed Solution

$\begin{array}{l} V = 36 kmph = 10 m / s \\ R = 100 \\ a_{c} = \frac{V^{2}}{R} = \frac{10^{2}}{100} = 1 \\ a_{Net} = \sqrt{a_{C}^{2} + a_{T}^{2}} \\ = \sqrt{{(\frac{V^{2}}{R})}^{2} + (0 .75)^{2}} \\ = \sqrt{1^{2} + \frac{3^{2}}{4^{2}}} = \frac{5}{4} m / s^{2} \end{array}$
$\begin{array}{l} a_{C} = a_{Net} \cos α \\ 1 = \frac{5}{4} \cos α \Rightarrow \cos α = \frac{4}{5} \\ α = 37^{\circ} \\ ∴ θ = 90 + α = 90 + 37 = 127^{\circ} \end{array}$