Explanation Given: The radius of curvature and velocity is shown in Figure (1). From Figure (1) the radius of curvature ρ is 40 m . The velocity υ of the boat is 0.0625 t 2 m/s . The Time t is 10 s . From Figure (1) write the expression for the velocity. υ = 0.0625 t 2 m/s (I) Here, velocity is υ . Write the expression for the instantaneous acceleration. a t = d υ d t (II). Here, tangential acceleration is a t and rate of change of velocity with respect to time ( t ) is d υ d t . Write the expression for normal acceleration. a n = υ 2 ρ (III). Here, normal acceleration is a n , constant velocity is υ and radius of curvature is ρ . Write the expression for the magnitude of acceleration. a = a t 2 + a n 2 (IV). Here, acceleration is a , tangential acceleration is a t and normal acceleration is a n . Conclusion: Substitute 0.0625 t 2 m/s for υ in Equation (II). a t = d υ d t = d d t ( 0.0625 t 2 m/s ) = 0.125 t m/s 2 (V)

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ENGR.MECH.:STAT.+DYNAMICS

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The magnitude of acceleration when t is 10 s .

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