A car starts at point P from rest, but with a constant tangential acceleration of 2 m / s 2 . The car is constrained to travel on a circular track of unknown radius. When the car reaches point Q , the

Question

A car starts at point P from rest, but with a constant tangential acceleration of $2 m / s^{2}$ . The car is constrained to travel on a circular track of unknown radius. When the car reaches point Q , the angle between the acceleration and velocity vectors is $3 7^{\circ}$ . The centre of the track is point O . What is the angle (in radians) POQ? If your answer is in the form of $α / β$ where $α & β$ are least integer values then find the value of $β - α$ .

TestHub · Accepted Answer

At Q, angle between velocity & acceleration is 37° as shown in figure.

$tan 3 7^{\circ} = \frac{a _{c}}{a _{t}}$

$a_{t} = 2 m / s^{2} ($ given $) \Rightarrow \frac{3}{4} = \frac{a _{c}}{2} \Rightarrow a_{c} = \frac{3}{2} m / s^{2}$

Let ' $ω$ ' be the angular speed of particle at ' $Q_{1}^{'}$ and ' $α$ ' be the angular acceleration $a_{t} = R α \Rightarrow R α = 2 \Rightarrow α = \frac{2}{R} rad / s^{2}$

$a_{c} = ω^{2} R \Rightarrow \frac{3}{2} = ω^{2} R \Rightarrow ω^{2} = \frac{3}{2 R} [$ At Q $]$

Now $ω_{f}^{2} = ω_{i}^{2} + 2 α Δ θ [ω_{i} = 0]$

$\frac{3}{2 R} = 0 + 2 \times \frac{2}{R} Δ θ$

$Δ θ = ϕ = \frac{3}{8} rad$

$\frac{3}{8} = \frac{α}{β} \Rightarrow β - α = 8 - 3 = 5$ Ans.

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