A car moves along an x axis through a distance of 965 m, starting at rest (at x = 0) and ending at rest (at x = 965 m). Through the first 1/4 of that distance, its acceleration is +2.10 m/s. Through the next 3/4 of that distance, its acceleration is -0.700 m/s. (a) What is its travel time through the 965 m? S (b) What is its maximum speed? m/s (c) Graph position x, velocity v, and acceleration a versus time t for the trip. (Do this on paper. Your instructor may ask you to turn in this work.) Constant-acceleration equations. Split the motion into two stages. During stage 1, you know the displacement, the initial velocity, and the acceleration, and you want the time. Then you want the velocity at the end of stage 1. 4. [0/1 Points] DETAILS PREVIOUS ANSWERS HRW9 2.P.069. How far does the runner whose velocity-time graph is shown below travel in 12 s? 76 x m Velocity (m/s) 8 + 1 1 1 0 4 8 12 16 Time (s) Distance is equal to the integration of velocity with respect to time. Here the integration can be done graphically. Math Help Algebra-Common Denominator, Algebra-Distribute Common Cross, Algebra-Quadratic Equation, Algebra-Squareroots & Squares Submit Answer MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER