A particle starts with angular acceleration $2 r a d / s e c^{2}$ . It moves $100 r a d$ in a random interval of $5 s e c$ . Find out the time at which random interval starts.

7.5 s e c

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4.5 s e c

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5 s e c

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6 s e c

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Solution

The correct option is A $7.5 s e c$
Starting from rest,
$θ_{2} = 0 + \frac{1}{2} \times 2 \times {(t_{2})}^{2} a n d θ_{1} = 0 + \frac{1}{2} \times 2 \times {(t_{1})}^{2} ∴ θ_{2} - θ_{1} = {(t_{2})}^{2} - {(t_{1})}^{2} 100 = (t_{2} - t_{1}) (t_{2} + t_{1}) ∴ t_{2} + t_{1} = \frac{100}{5} = 20 n o w, t_{2} + t_{1} = 20 a n d t_{2} - t_{1} = 5 \Rightarrow t_{1} = 7.5 s$

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A particle starts with angular acceleration 2 rad/sec2. It moves 100 rad in a random interval of 5 sec. Find out the time at which random interval starts.

The correct option is A 7.5 secStarting from rest,θ2=0+12×2×(t2)2andθ1=0+12×2×(t1)2∴θ2−θ1=(t2)2−(t1)2100=(t2−t1)(t2+t1)∴t2+t1=1005=20now,t2+t1=20andt2−t1=5⇒t1=7.5s

A particle starts with angular acceleration $2 r a d / s e c^{2}$ . It moves $100 r a d$ in a random interval of $5 s e c$ . Find out the time at which random interval starts.