## Find Acceleration Diagram from Velocity Vector Diagram| University Physics

### The figure above shows the motion diagram for an object. If the x-axis is chosen as shown (positive to the right), which of the diagrams below best represents the corresponding acceleration vs time graph? (In the graph, later time is to the right). ## Pendulum’s acceleration at the lowest point| University Physics

### In a physics lab, a pendulum hung from the ceiling slowly swings back and forth. Select the arrow that best indicates the direction of the pendulum’s acceleration when it reaches the lowest point moving from left to right. ## Rank the diagrams according to the x-component of the acceleration| University Physics

### The figure below shows six different motion diagrams. The separation between dots is in the same scale in all diagrams, and the time interval between dots is also the same for all of them. The direction of the positive x-axis is also shown. Rank the diagrams according to the x-component of the acceleration. (Positive is greater than negative) ## Finding where the magnitude of acceleration is constant from given motion diagrams| University Physics

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### The main graph shows position versus time for a certain object. Which of the graphs below correctly depicts the object’s acceleration?  ## Draw the acceleration-time graph given a parabolic position-time graph| University Physics

### The main graph shows position versus time for a certain object. Which of the graphs below correctly depicts the object’s acceleration?  ## Solution:

### Part A

We are given $v_0=22.0\:m/s$ $a=-0.150\:m/s^2$ $x=210\:m$

We are required to solve for time, t. So we have $x=v_0t+\frac{1}{2}at^2$ $\left(210\:m\right)=\left(22.0\:m/s\right)t+\frac{1}{2}\left(-0.150\:m/s^2\right)t^2$

If we rearrange the terms into a general quadratic equation, we have $0.075t^2-22t+210=0$

Solve for t using the quadratic formula. We are given $a=0.075;\:b=-22t;\:c=210$ $\displaystyle t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle t=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}$ $t=9.88\:s\:\text{and}\:t=283.46\:s$

Discard t=283.46 s as it can be seen from the problem that this is not a feasible solution. So, we have $t=9.88\:s$

### Part B

We have the same given values from part a. So we have $v^2=v_0^2+2ax$ $v=\sqrt{v_0^2+2ax}$ $v=\sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(210\:m\right)}$ $v=20.6\:m/s$

### Part C

We are given the same values as in the previous parts, except that for the value of x since we should incorporate the length of the train. For the distance, x, we have $x=210\:m+130\:m$ $x=340\:m$

Therefore, we have $x=v_ot+\frac{1}{2}at^2$

You can solve this problem by the method presented in part A. But we can also rearrange the equation to solve for t, using also quadratic equation. We have $\displaystyle t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}$ $\displaystyle t=\frac{-\left(22.0\:m/s\right)\pm \sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(340\:m\right)}}{-0.150\:m/s^2}$ $t=16.4\:s$

### Part D

We have the same given values. $v^2=v_0^2+2ax$ $v=\sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(340\:m\right)}$ $v=19.5\:m/s$