The velocity of a particle as a function of time is given by the graph below: velocity vs time 20 15 10 -10 -15 time (s) Part A The displacement of the particle would best be approximated as O -5 m O -20 m O 20 m O 95 m

You qualitatively analyzed the motion of a van earlier. Now, using the example of the ball thrown into the air, you can do a more detailed analysis of the van's motion. The table shown here includes the time and position data, with one worked example for finding acceleration. Time Position Velocity Ad Acceleration t(s) (m/s) Ar (m/s") 0.0 0.0 6.0 2.0 12 +3.0 12 4.0 36 Sample Calculation Notice that the velocity that will be plotted at t = 1.0 s is the average velocity between t = 0.0 s and t 2.0 s. The velocity that will be plotted at t 3.0 s is the average velocity between t = 2.0 s and 4.0 s. The acceleration that will be plotted at t= 2.0 s is the average acceleration between t = 1.0 s and t = 3.0s. 6.0 48 8.0 96 10.0 142 12.0 190 14.0 226 16.0 250 Adgd-d 12 m-0.0 m 2.0 s-0.0 s V = 18.0 262 Alo2 t2-lo 12 m 2.0 s %3D Analyze and Conclude 1. How well do the average and instantaneous velocities that you calculated agree with = 6.0 m each other? 36 m-12 m V= Afz4 %3D %3D 2 Separate the…

The following graph depicts the velocity of someone playing tag over a short period of time. Use this velocity graph to answer the following questions. The person's maximum magnitude for velocity is 4.5 m/s in both the positive and negative directions.  A is at time point 3 seconds, B is at time point 4.5 seconds, C is at time point 5.5 seconds, D is at time point 7 seconds, E is at 8 seconds, F is at 9 seconds and G is at 10 seconds. What is the acceleration in meters/second2 (m/s2) at the 2 second mark of playing tag? What is the acceleration in meters/second2 (m/s2) at the 5 second mark?  What is the velocity in meters/second (m/s) at the 5 second mark?  What is the acceleration in meters/second2 (m/s2) at the 6.5 second mark?  What is the acceleration in meters/second2 (m/s2) at the 8.5 second mark?  What is the velocity in meters/second (m/s) at the 9.5 second mark?

PLEASE ANSWER LETTERS B TO DThe position of a particle is given by the equation s(t) = t3 − 6t2 + 9t, where t is in seconds and s in meters.a) What is the velocity after 2 s? After 4 s?b) Find the instant when the particle reverses directionc) Determine the time intervals when it is moving to the LEFT? RIGHT?d) Find the initial position and the positions when it changes direction

The slope of the graph is determined to be 0. What does this slope determine about the motion?

Short computation The velocity of an object as a function of time is given by v(t) = b = ct- et²2, where v and t are in SI units. The other variables are given as b = 4.1 m/s, c = 4.1 m/s², and e = 1.5 m/s³. Determine the instantaneous acceleration of the object at time t = 0.38 s. a m/s² Remember to include a "-" if the acceleration turns out to be negative. Do not include a "+" for a positive value. Report your numerical answer below, assuming three significant figures.

This graph shows the position of a particle as a function of time. What is its instantaneous velocity at t = 7 s? x (m), 12 1(s) 6. 2 O 3 m/s -3 m/s O 12 m/s O -12 m/s O Need additional information

Help, please

The positions of objects A and B with respect to Earth depend on time as follows: x (t) = 10.0m (4.0m/s)t and x₂(t) = 12m + (6m/s)t. Represent their motions using position-time graphs. Use two different methods to find where the two objects will meet.

Please help with question 3. Thank you.