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Grade: 12
        
please..................................................................
one month ago

Answers : (1)

Arun
22299 Points
							
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
 
one month ago
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