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An object undergoes acceleration of 2.3(ihate + 3.6(jhat)m/s^2 over a 10s interv

ID: 1752876 • Letter: A

Question

An object undergoes acceleration of 2.3(ihate + 3.6(jhat)m/s^2 over a 10s interval. At the end of this time, its velocity is33(ihat) + 15(jhat) m/s. A) What is it's velocity at the beggining of the 10-sinterval B) By how much did the speed change? C) By how much did its direction change? D) Show that the speed change is not given by the magnitude ofthe acceleration multiplied by time? Why is it not? An object undergoes acceleration of 2.3(ihate + 3.6(jhat)m/s^2 over a 10s interval. At the end of this time, its velocity is33(ihat) + 15(jhat) m/s. A) What is it's velocity at the beggining of the 10-sinterval B) By how much did the speed change? C) By how much did its direction change? D) Show that the speed change is not given by the magnitude ofthe acceleration multiplied by time? Why is it not?

Explanation / Answer

PART A>>>>>    defn     a = (vf - vo) /t,     so      vo = vf - a t   where bothvf and a are given in the problem statement. thus,    vo = 33 i + 15 j - 23i - 36j   =   10 i - 21 j    this isthe initial velocity vector at the beginning of the 10 secinterval. PART B>>>>   Initial speed is magnitudeof vo , which is (102+212 ) = 23.26 m/s Final speed is magnitude of vf , which is (332 + 152 ) =   36.25m/s change in speed is final minus inititial, which is 36.25- 23.26 =   12.99 m/s PART C>>>>>   direction of initialvelocity vector is   invtan (- 21/10) = -64.5 deg direction of final velocithy is     invtan(15/33) = 24.4deg         change in direction is 64.5 + 24.4   = 88.9 degrees. PART D>>>>>   magnitude of a is (2.32 + 3.62 ) = 4.27 m/s2    a t   = ( 4.27) (10s ) =   42.7 m/s   and this is NOT close to thespeed change of 12.99 m/s found previously. the reason these two results are not equal is because BOTHspeed and direction are changing at the same time in this problem,and the simple eqn of speed change = at    is ONLY true for straight line motion at aconstant acceleration.
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