Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

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When acceleration is constant, the average velocity is given by

where
and
are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

where
are the final/initial displacements, and
are the final/initial times, respectively.
Take the car's starting position to be at
. Then

So we have

You also could have first found the acceleration using the equation

then solve for
via

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of
anyway.
We are given the gravitational potential energy and the height of the ball and is asked in the problem to determine the mass of the ball. the formula to be followed is PE = mgh where g is the gravitational acceleration equal to 9.81 m/s^2. substituting, 58.8 J = m*9.8 m/s^2 * 30 m; m = 0.2 kg.
Average speed =
(distance covered during some period of time)
divided by
(length of time to cover that distance).
The area under the velocity time graph is 125 m and the meaning of the area is displacement.
<h3>
What is area under velocity - time graph?</h3>
The area under a velocity time graph represents the displacement of the object.
total area of the graph = A1 + A2
total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)
total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)
total area of the graph = 125 m
Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.
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