6: Short way: it cannot be 2.5, 3, or 5 because up to 5 seconds it only has positive velocity so it must be moving forwards.
Long Way: Velocity is in m / s, multiply that by time (s) to get m or displacement. From 0->5 you have a triangle under the curve, (1/2)(5)(20) = 50 meters displaced positive, you need to then look when velocity is under the curve and use a similar equation to solve for the area but make the answer negative. Find the point where it equals -50 and that is where it will have returned.
Answer to 6: B
7. I cannot see the problem enough to answer this. Just know if the line is above 0 velocity is positive so it is moving the direction it started, when it goes below 0 velocity is negative so it is moving opposite direction it started.
8. Accelration is change in velocity. Whatever the slope of the velocity graph is acceleration. At t=8 the slope is 0 because it is not going up or down.
Answer to 8: A
Complementary angles are those whose sum is 90 degrees, right?, then let's find A, A=72 degrees, because its mid point is 36 deg. Then, B = 18 degrees and its mid point is 9 degrees! :)
Most likely answer would be A
Answer:
Part a)

so here the angle made by the string is independent of the mass
Part b)

Explanation:
Part a)
Let the string makes some angle with the vertical so we have force equation given as


so we will have


so here the angle made by the string is independent of the mass
Part b)
Now from above equation if we know that angle made by the string is

so we will have




The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
Learn more here: brainly.com/question/24430414