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# Graphical Kinematics: Velocity vs. Time

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## Graphical Kinematics: Velocity vs. Time Answer Key

Instructions: An object's changing velocity is depicted in the Velocity vs. Time Graph below. Use the graph to answer the following questions. Descriptions must be written in complete sentences. For any questions involving calculations, you must show ALL your computation work. Acceleration answers should be in $m/(s^2)$ units.

1.
Describe, in detail, the motion of the object during the time interval from A to B on the graph.
• The object is speeding up in a positive direction from an initial velocity of zero at point A to a velocity of 20m/s at point B. The straight upward sloping line indicates that the change in velocity from A to B is constant or Uniform Accelerated Motion.
2.
Calculate the acceleration of the object during the time interval from A to B.
• $a=(Deltav)/(Deltat)=(v_B-v_A)/(t_B-t_A)=(20m/s-0m/s)/(2.0 s-0s)=(20m/s)/(2s)=10m/(s^2)$
3.
Determine the change in position of the object during the time interval from A to B.
• $x=x_A+v_At+(1/2)at^2$
$x=0+(0*2s)+(1/2)(10m/(s^2))(2s)^2=10m$
4.
Describe, in detail, the motion of the object during the time interval from B to C.
• The object is moving at a constant velocity of 20m/s for two seconds from B to C. Because there is no change in velocity, acceleration is zero during this time interval from B to C.
5.
Determine the change in position of the object during the time interval from B to C.
• $x=x_B+v_Bt+(1/2)at^2=10m+20m/s*2s+(1/2)(0m/(s^2))(2s)^2=50m$
So from B to C, the change in position is 50m-10m=+40m.
6.
Describe, in detail, the motion of the object during the time interval from C to D.
• The object is speeding up in a positive direction from an initial velocity of 20m/s at point C to a velocity of 40m/s at point D. The straight upward sloping line indicates that the change in velocity from C to D is constant or Uniform Accelerated Motion.
7.
Calculate the acceleration of the object during the time interval from C to D.
• $a=(v_D-v_C)/(t_D-t_C)=(40m/s-20m/s)/(5s-4s)=20m/(s^2)$
8.
Determine the change in position of the object during the time interval from C to D and how far it has moved in total from time=0.
• $x=x_C+v_Ct+(1/2)at^2$
$x=x_C+v_Ct+(1/2)(at^2)=50m+(20m/s*1s)+(1/2)(20m/(s^2))(1s)^2=80m$
So, from time zero, the object has moved 80m, and the change in position from C to D = 80m-50m=30m.
9.
Describe, in detail, the motion of the object during the time interval from E to F.
• The object is speeding up in a negative direction from an initial velocity of 40m/s at point E to a velocity of 20m/s at point F. The straight downward sloping line indicates that the change in velocity from E to F is constant or Uniform Accelerated Motion.
10.
Calculate the acceleration of the object during the time interval from E to F, and the final position of the object at H.
• $a=(v_F-v_E)/(t_F-t_E)=(20m/s-40m/s)/(10s-8s)=(-20m/s)/(2s)=-10m/s^2$
The acceleration during the interval from E to F = $-10m/s^2$.
$x_E=x_D+v_Dt+(1/2)at^2=80m+(40m/s*3s)+(1/2)(0m/(s^2))(3s)^2=80m+120m=200m$
$x_F=x_E+v_Et+(1/2)at^2=200m+(40m/s*2s)+(1/2)(-20m/(s^2))(2s)^2=240m$
$x_G=x_F+v_Ft+(1/2)at^2=240m+(20m/s*1s)+(1/2)(0m/(s^2))(1s)^2=260m$
$a_"G-H"=(30m/s-20m/s)/(1s)=+10m/(s^2)$
$x_H=x_G+v_Gt+(1/2)at^2=260m+(20m/s*1s)+(1/2)(10m/(s^2))(1s)^2=285m$
+285m from the original position A is the final position of the object at H.
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