Question

The motorcycle is traveling at 8 m/s when it is at A. The speed is then increased at v? = 0.2 m/s2 .

b: determine the acceleration of the motercycle at the instant t = 7 s

Problem 12.123 Part A The motorcycle is traveling at 8 m/s when it is at A. The speed is then increased at Determine the speed of the motorcycle at the instant t7 s Express your answer to three significant figures and include the appropriate units. 7 s. = 0.2 m/s2 . (Figure 1) u= 9.40 Submit My Answers Give Up Correct Part B Determine the acceleration of the motorcycle at the instant t7 s Express your answer to three significant figures and include the appropriate units. aValue Units Submit My Answers Give Up Incorrect; Try Again; 2 attempts remaining Figure 1 of 1 y-0.5x2

Answer 1

The equation for the path travelled by the motor cycle is, y = 0.5 x^2 dy/dx = 0.5(2x) = x d^2 y/dx^2 = 1 The radius of the acceleration is, rho = squareroot (1 + (dy/dx)^2)^3/(d^2 y/dx^2) = squareroot (1 + x^2)^3/1 = (1 + x^2)^3/2 From the equation of the kinematics: the speed of the motorcycle is v = v_0 + v t = 8 m/s + (0.2 m/s^2) (7s) = 9.4 m/s The distance travelled by the motorcycle is, s = v_0 t + 1/2 v t^2 = (8) (7) + 1/2 (0.2) (7)^2 = 60.9 m The expression for the distance is, s = integrals^x_0 squareroot 1 + (dy/dx)^2 dx 60.9 = integrals^x_0 squareroot 1 + x^2 dx (say x = (dy/dx)) = 1/2 (x squareroot x^2 + 1 + sin h^-1 x) (x squareroot x^2 + 1 + sin h^-1 x) = 121.8 The radius of the acceleration becomes, rho = (1 + x^2)^3/2 = (1 + (10.8)^2)^3/2 = 1275.95 m The normal acceleration is, a_normal = v^2/rho = (9.4)^2/1275.95 = 0.069 m/s^2..