Consider the timedependent velocity of a car shown in the fi

Consider the time-dependent, velocity of a car shown in the figure below. Calculate the acceleration a(t) for 0 le t le 16 except for at times t = 2,10,12. Why are these points problematic? Use the Fundamental Theorem of Calculus (given below), to compute the position function x(t) for 0 le t le 16. The Fundamental Theorem tells us that for any a R, x(t) - x(a) = integra^b_a v(s) ds, when x\'(s) = v(s), thus x(t) = integral^b_a v(s) ds + x(a) Use the initial condition, x(0) = 0 and the fact that position, x(t) must be a continuous function to find the constants of integration. Answers which do not show work will receive zero credit. Next, graph x[t) for 0 le t le 16. See the \"Labs\" page in Canvas for a MATLAB file to help you plot x{t). You aren\'t required to use MATLAB, but you must use a computer tool to create the plot. Sage and Maple are also good options. Staple your printed plot at the end of this lab. You do not need to include any code you used to generate the plot. Use your equation for x(t) to determine the total distance the car has traveled at t = 11 seconds. Be careful: x(t) is the equation for the position of the car at any time t; it gives the displacement of the car from its initial position. If the velocity ever changes sign then the position (displacement) probably won\'t match the total distance traveled because we count travel in both the positive and negative directions as positive distance traveled when measuring total distance.

Solution

(a) Acceleration is defined as the derivative of the velocity, since at the point t=2,10 and 12, there is a spike which means the derivative doesn\'t exists at these points, hence we can\'t find the acceleration at these points

Acceleration from t belonging to [0,2)

a = [v(2)-v(0)]/(2-0) = 8/2 = 4 m/s^2

Acceleration from t belonging to (2,10)

a = [v(10)-v(2)]/(10-2) = 0 m/s^2

Acceleration from t belonging to (10,12)

a = [v(12)-v(10)]/(12-10) = -2 m/s^2

Acceleration from t belonging to (12,16]

a = [v(16)-v(12)]/(16-12) = 0 m/s^2

c) The distance x(t) will be the area under the graph from t=0 to t=11 seconds

x(t) = Area under the v-t graph

=> 1/2 * 2 * 8 + 8 * 8 + 6 + 1

=> 8 + 64 + 7

=> 79 m