A particle\'s acceleration is described by the function ax=10-t m/s^2 , where t

Question

A particle's acceleration is described by the function ax=10-t m/s^2 , where t is in s. Its initial conditions are Xo = 260 and Vox = 0 m/s at t=0 . a.)At what time is the velocity again zero? Express your answer with the appropriate units. b.)What is the particle's position at that time? Express your answer with the appropriate units.

Explanation / Answer

It appears that we are dealing with linear motion of a particle, therefore I will ommit the subcript x in ax and vox for simplicity. Acceleration: a = 10 - t Velocity is an integral of aceleration: v = Integral a(t) dt v = 10t - 1/2 t² + C integral constant C is defined by initial conditions: t = 0, v= 0, 0 = 10*0 + 1/2*0² + C -----> C = 0 , and v = (10 - 1/2 t) t At what time is v again 0? v = 0 when t= 0 (above said) and when 10 - 1/2 t = 0, or t = 20 seconds. What is the particle's position at that time (t = 20 s)? Coordinate x is an integral of v. x = Integral v(t) dt = Integral (10 t - 1/2 t² ) dt x = 5 t^2 - 1/6 t^3 + C, C is defined by initial conditions: t = 0, x = 0 ---> C = 0 x = 5t^2 - 1/6 t^3 When t= 20, x = 5 * 20^2 - (1/6) * 20^3 = 5 * 400 - (1/6) * 8000 x = 2000 - 1333 x = 667 m

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