The velocity of a particle moving on the x-axis is given by v=x^2+x, where x is in m and v is in m/s. What is its position when its acceleration is 30 m/s^2

Asked by Anil | 18th Jun, 2017, 07:59: PM

Expert Answer:

begin mathsize 12px style Velocity space of space the space particle space is
straight v open parentheses straight x close parentheses equals straight x squared plus straight x
Acceleration space is
straight a equals dv over dt equals dv over dx cross times dx over dt equals straight v dv over dx space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis since space dx over dt equals straight v right parenthesis
therefore straight a equals open parentheses straight x squared plus straight x close parentheses straight d over dx open parentheses straight x squared plus straight x close parentheses equals open parentheses straight x squared plus straight x close parentheses open parentheses 2 straight x plus 1 close parentheses
therefore straight a equals 2 straight x cubed plus straight x squared plus 2 straight x squared plus straight x equals 30 space left parenthesis given right parenthesis
therefore 2 straight x cubed plus 3 straight x squared plus straight x equals 30
Solving space this space we space get space straight x equals 2 space straight m end style

Answered by Romal Bhansali | 1st Nov, 2017, 01:54: PM

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