Given the equation for position of a car moving at constant velocity v_x (t): x(

Question

Given the equation for position of a car moving at constant velocity v_x (t): x(t) = integral_0^t v dt This is called a DEFINITE integral, because the initial and final times are known. Here t = 0 seconds, and t_F is some final time. Here is the solution: x(t) = integral_0^t v_x dt = (v_x t) = v_x middot t_F - v_x middot 0 = v_x t_F The Definite Integral is described as the AREA UNDER THE CURVE, when the function is plotted on the vertical axis and the variable is plotted on the horizontal axis. Let the velocity be a constant, V_x = 22m/s. Here is the plot of velocity as a function of time: Here is the same plot, but for a DEFINITE time range from Here t_1 = 0 seconds, and t_F = 2 seconds. The AREA shown is [22m/s] middot [2s] = 44 meters the position of the car at t_F = 2 seconds.

Explanation / Answer

As I can see liitle bit from the image, the question asked is: Calculate the area under the curve for t=6 seconds.

Ans: As veocity is constant throughout the motion v=22m/s, therefore area=22*6=132meters which is also the position of the body after 6 seconds.

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