The acceleration, ax = 5 m/s3 t + 2 m/s The velocity, in the x direction, is given by the formula: vx = 2.5 m/s + 2 m/s t (A) What is the acceleration at t = 6.1 s? a(t=6.1 s) = m/s (B) What is the area under the acceleration graph between t= 0s and t = 6.1 s? m/s. Consider the graph below. Hint: The area under consideration forms a triangle with a height of (5 m/s3 x 6.1 s) and a width of 6.1 s on top of a rectangle with a height of 2 m/s2 and a width of 6.1 s. Alternately, try the trapezoid formula for area under a function: Area = (y + y)(x-x) = [a (0 s) +a (6.1 s)](6.1 s - 0 s) (C) What is the change in velocity between t= 0s and t = 6.1 s? Avxvx(final) (initial) = v(6.1 s) - v(0 s) = Acceleration (m/s^2) in the x direction, is given by the formula: 50 40 30 201 10 Time (s) m/s