please..................................................................
please..................................................................
Abhijeet Nath , 6 Years ago
Grade 12
Mechanics> please......................................
1 Answers
Arun
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
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