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6 months ago

Arun
23315 Points
							Given, distance, s = t² + 5differentiate with respect to time,ds/dt = speed ,v = 2tat t = 5√(3/10) , speed , v = 2 × [5√(3/10)] = 10√(3/10) m/sso, 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/s²now rate of change of speed, dv/dt = 2e.g., tangential acceleration , a_tat​ = 2 m/s²now, 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/s²given, distance, s = t² + 5differentiate with respect to time,ds/dt = speed ,v = 2tat t = 5√(3/10) , speed , v = 2 × [5√(3/10)] = 10√(3/10) m/sso, 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/s²now rate of change of speed, dv/dt = 2e.g., tangential acceleration , a_tat​ = 2 m/s²now, 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

6 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions