Question

Using these two equations (behr free fall exp)

x = -0.20t^2 + 0.60t + .03 and V= 9.70t + 0.9625

find the acceration of the fall and the velocity at “t=0” for both equations. Then calculate the velocity at time t=0.2 for each equation. (I think you use derivitives for the second part)(please show work)

	acceleration	Initial velocity	Velocity at t=0.2
v vs t
x vs time
Average acceleration

Answer 1

For x vs t, equation
x = -0.2 t² + 0.6 t + 0.3
v = dx/dt = - 0.4 t + 0.6
Hence initial velocity v(t=0) = 0.6 and
   v(t=0.2) = -0.4(0.2) + 0.6 = 0.52

a = dv/dt = -0.4
Hence acceleration for this equation is -0.4

For equation
v = 9.70t + 0.9625
Initial velocity v(t=0) = 0.96
   v(t=0.2) = 9.7(0.2) + 0.9625 = 2.90

a = dv/dt = 9.70