Given:
∣ac∣=∣at∣
⇒rv2=dtdv
⇒∫4vv2dv=∫0trdt
⇒[v−1]4v=rt
⇒v−1+41=2t
⇒v=1−8t4=dtds
⇒4∫0t1−8tdt=∫0sds
(r=0.5m,distancecovereds=2πr=π)
⇒4×−8[ln(1−8t)]0t=π
⇒ℓn(1−8t)=−2π
⇒1−8t=e−2π
⇒t=(1−e−2π)81s
So, α=8.
A particle is moving in a circle of radius 50cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is 4ms−1, the time taken to complete the first revolution will be α1[1−e−2π]s, where α=______.
Held on 29 Jan 2024 · Verified 6 Jul 2026.
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