Question

2D Position, Velocity, and Acceleration Relationships with Calculus The velocity as a function of time for an asteroid in the asteroid belt is given by e-t/to where vo and to are constants Use ti 0 as the initial time, and tf -1.3to as the final time. The values for the constants that you will use are: vo -39 m/s to 630 s Find the displacement of the asteroid. 1pts Submit Answer Incorrect. Tries 8/20 Previous Tries Find the x component of the acceleration at the final time ar (tt) 1pts Submit Answer Incorrect. Tries 1/20 Previous Tries Find the magnitude of the acceleration at the final time. a(tt) 1pts Submit Answer Tries 0/20

Answer 1

V(t) = V_0 e^-t/t)_0 t^^+ v_0 t/2 t_0 j^^ = v_x i^^+ v_y j^^ (a) delta r^vector = integral V_x dt i^^+ integral v_y dt j^^ = integral^1.3 t_0_0 V_0 e^-t/t_0 dt i^^+ integral^1.3 t_0_0 v_0 t/2 t_0 dt j^^ = (-v_0 t_0 e^-t/t_0)^1.3 t_0_0 i^^+ (v_0 t/2 t_0 dt j^^ = -v_0 t_0 e^-t/t_0)^1.3 t_0_0 i^^+ (v_0/2 t_0 t^2/2)^1.3 t_0_0 = -v_0 t_0 (e^-1.3 - 1) i^^+ (1.3)^2 v_0 t_0/ j^^ = 1.78 times 10^4 i^^+ 1.04 times 10^4 j^^ (b) a_x = dv_x/dt = d/dt (v_0 e^-t/t_0) = -v_0/t_0 e^-t/t_0 = -39/630 times e^-1.3 = -1.7 times 10^-2 m/s^2 (c) Vertical-bar a Vertical-bar = squareroot a_x^2 + a_y^2 a_y = dv_x/dt = d/dt (v_0 t/2t_0) = v_0/2 t_0 = 39/2 times 630 = 3.09 times 10^-2 m/s^2 a = squareroot (-1.7 times 10^2)^2 + (3.09 times 10^-2)^2 = 3.5 times 10^-2 m/s^2 (d) (V_x)_average = 1/T integral^