Question

A point P moves in counter clockwise direction on a circular path.The movement of P is such that it sweeps out a length s =t^2 + 5, where s is in metres and t in sec.The radius of the path is 20m.The accln of P when t=5√(3/10) sec is nearly??





plzz ans!!

Answer 1

given, distance, s = t² + 5

differentiate with respect to time,

ds/dt = speed ,v = 2t

at t = 5√(3/10) , speed , v = 2 × [5√(3/10)] = 10√(3/10) m/s

so, centripetal acceleration, $a_c$ = v²/R, where R is the radius of the path. so, $a_c$ = [10√(3/10)]²/20 = 30/20 = 1.5 m/s²

now rate of change of speed, dv/dt = 2

e.g., tangential acceleration , $a_t$ = 2 m/s²

now, net acceleration, $a_{net}$ = $\sqrt{a_c^2+a_t^2}$

= $\sqrt{1.5^2+2^2}$ = 2.5 m/s²

Answer 2

given, distance, s = t² + 5

differentiate with respect to time,

ds/dt = speed ,v = 2t

at t = 5√(3/10) , speed , v = 2 × [5√(3/10)] = 10√(3/10) m/s

so, centripetal acceleration, a_cac​ = v²/R, where R is the radius of the path. so, a_cac​= [10√(3/10)]²/20 = 30/20 = 1.5 m/s²

now rate of change of speed, dv/dt = 2

e.g., tangential acceleration , a_tat​ = 2 m/s²


now, net acceleration, a_{net}anet​ = \sqrt{a_c^2+a_t^2}ac2​+at2​​

= \sqrt{1.5^2+2^2}1.52+22​ = 2.5 m/s²


given, distance, s = t² + 5

differentiate with respect to time,

ds/dt = speed ,v = 2t

at t = 5√(3/10) , speed , v = 2 × [5√(3/10)] = 10√(3/10) m/s

so, centripetal acceleration, a_cac​ = v²/R, where R is the radius of the path. so, a_cac​= [10√(3/10)]²/20 = 30/20 = 1.5 m/s²

now rate of change of speed, dv/dt = 2

e.g., tangential acceleration , a_tat​ = 2 m/s²


now, net acceleration, a_{net}anet​ = \sqrt{a_c^2+a_t^2}ac2​+at2​​

= \sqrt{1.5^2+2^2}1.52+22​ = 2.5 m/s²