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If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distance it travels each succeeding second would be:_________
a) constant.
b) less and less each second.
c) greater than the second before

Respuesta :

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

[tex]h=ut+\frac{1}{2}gt^2[/tex]

here u=0 and t=time taken

[tex]h=\frac{1}{2}gt^2[/tex]

for [tex]t=1 s[/tex]

[tex]h_1=\frac{1}{2}g[/tex]

for [tex]t=2 s[/tex]

[tex]h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g[/tex]

distance traveled in 2 nd sec[tex]=2g-\frac{1}{2}g=\frac{3}{2}g[/tex]

for [tex]t=3 s[/tex]

[tex]h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g[/tex]

distance traveled in 3 rd sec[tex]=\frac{9}{2}g-2g=\frac{5}{2}g[/tex]

so we can see that distance traveled in each successive second is increasing

The amount of distance travels by free falling object in each succeeding second is greater than the second before.

What is the speed of free falling body?

Speed of the free falling object is the speed, by which the body is falling downward towards the ground level.

At the Earth's surface, the speed of the free falling object will accelerate 9.841 meters per squared second.The odometer is the device, which is used the speed of the body in m/s or km/hrs.

Now we have to find out, whether the distance traveled by the object each succeeding second is constant, deceasing or increasing.

The second equation of the motion for distance can be given as,

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Here,  [tex]u[/tex] is the initial velocity of the body, [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by it.

As for the free falling body the initial velocity is zero. Thus the distance traveled by it for the first second is,

[tex]s=0+\dfrac{1}{2}9.81\times1\\s=4.905\rm m[/tex]

The distance traveled by it for the 2nd second is,

[tex]s=\dfrac{1}{2}9.81\times2^2\\s=19.61\rm m[/tex]

The distance traveled by it for the 3rd second is,

[tex]s=\dfrac{1}{2}9.81\times3^2\\s=44.145\rm m[/tex]

Now the difference between the third and 2nd second of distance traveled by object is greater then the difference between the second and first second.

Thus, the amount of distance travels by free falling object in each succeeding second is greater than the second before.

Learn more about speed of free falling body here;

https://brainly.com/question/24520854

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