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(B) Given: Radius $R = 600\,m$,initial speed $u = 54\,km/h = 15\,m/s$.Tangential acceleration $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$.Centripetal accel

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Given: Radius $R = 600\,m$,initial speed $u = 54\,km/h = 15\,m/s$.Tangential acceleration $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$.Centripetal acceleration $a_c = \frac{v^2}{R}$.Given $a_t = a_c$,so $v\frac{dv}{ds} = \frac{v^2}{R} \Rightarrow \frac{dv}{v} = \frac{ds}{R}$.Integrating both sides: $\int_{15}^{v} \frac{dv}{v} = \int_{0}^{s} \frac{ds}{R} \Rightarrow \ln(\frac{v}{15}) = \frac{s}{R} \Rightarrow v = 15e^{s/R}$.Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = 15e^{s/R} \Rightarrow e^{-s/R} ds = 15 dt$.To find the time $T$ for the first quarter revolution,$s$ goes from $0$ to $\frac{\pi R}{2}$.$\int_{0}^{\pi R/2} e^{-s/R} ds = \int_{0}^{T} 15 dt$.$[-R e^{-s/R}]_{0}^{\pi R/2} = 15T$.$-R(e^{-\pi/2} - 1) = 15T \Rightarrow R(1 - e^{-\pi/2}) = 15T$.$T = \frac{600}{15}(1 - e^{-\pi/2}) = 40(1 - e^{-\pi/2})$.Comparing with $t(1 - e^{-\pi/2})$,we get $t = 40$.

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