Calculate the acceleration in the intervals B-C , and the position in point C an
ID: 2046974 • Letter: C
Question
Calculate the acceleration in the intervals B-C, and the position in point C and please show steps as to how you got your answer! Thank you! :)
Explanation / Answer
in the intervals B-C the curve is parabola. So, lets assume the equation of parabola is x= a* t^2 + b*t+ c the max or min point of a parabola happens at t= -b/ (2*a) from the graph - b/ (2a) = 4 =>b=-8a ---------------- (1) from A - B the equation of curve is x= 0.52 - 3*t^2 +1.19*t^3 at t=2 , x= -1.96 hence, the point (2,-1.96) also satisfies the parabola from B-C putting t=2 and x=-1.96 in the parabola we get 4a+2b+c = -1.96 putting the value of b from equation (1) and solving for c we get c=12a- 1.96 -------------(2) therefore the equation from B-C is x(t) = a*t^2 -8*a*t+12*a-1.96 now the slope B is same for both the curves comparing the first order derivative of both the curves at t=2 we get a=-0.57 the acceleration is defined as the second order derivative of distance time curve so, the acceleration from B-C is d^2 x/ d t^2= 2*a = 2*(-0.57) = - 1.14 m/s^2 position at point C = x(4) = -0.57* 16 + 8 *0.57*4 - 12*0.57-1.96 = 0.32m