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Len [333]
3 years ago
7

If the average speed of a person was 1.2 meters/second, does this mean that their speed was exactly 1.2 meters/second the whole

time? Is the average speed related to the maximum or minimum of the person? Explain why do you think so.
Physics
1 answer:
alexandr1967 [171]3 years ago
7 0

Average speed is defined by the following formula

v = \frac{D}{t}

here

D = total distance that an object move from its initial position to final position

t = total time of the motion

so here we will say that there is no such relation between initial or final speed or we can say maximum or minimum speed of object with average speed of object.

We only need to find the total distance and total time of motion in order to find the average speed

here we can see many examples like let say an object moves with speed v1 for time t1 and then with speed v2 for time t2 then here average speed is given as

d = v_1 t_1 + v_2t_2

since we know that distance covered is product of speed and time

that's why we used above equation for finding total distance

now the average speed will be

v_{avg} = \frac{v_1t_1 + v_2 t_2}{t_1 + t_2}

so this is how we can find the average speed for above motion

so average speed is always between maximum and minimum speed any value in-between.

It is neither the maximum value nor it is minimum value

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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

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