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sergejj [24]
3 years ago
15

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels

Physics
1 answer:
Aleksandr [31]3 years ago
3 0

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

v_f=v_o+at\qquad\qquad [1]

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

(c) From [2] we can solve for a:

\displaystyle a= 2\frac{x-v_ot}{t^2}

Since vo=8 m/s, x=640 m, t=40 s:

\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

v_f=8+0.4\cdot 40

v_f=8+16=24

The final velocity is 24 m/s

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The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

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