Question

A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward and away from the sensor, as shown above. The positive direction is defined as the direction away from the sensor. Which of the following describes the student’s final position xf in relation to the starting position x0 and the student’s average horizontal acceleration ax between 0.0 s and 3.0s? A.)Position xf is farther away from the sensor than x0, and ax is positive. B.)Position xf is farther away from the sensor than x0, and ax is negative. C.)Position xf is closer to the sensor than x0, and ax is positive. D.)Position xf is closer to the sensor than x0, and ax is negative.

Answer 1

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

                     Area of trapezoidal are=

                            =$\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)$

                            =$\frac{1}{2} *(2.25-1.75*0.5)$

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

                                  ax=(change in velocity)/(Time)

                                  ax=$\frac{-0.5-0}{3} =-\frac{1}{6} ms2$