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salantis [7]
3 years ago
10

A car accelerates from 45m/s to 0 m/s in 10 seconds what is the average acceleration of the car

Physics
1 answer:
jek_recluse [69]3 years ago
3 0

Answer: -4.5 m/s^2

Explanation:

The average acceleration of the car is given by:

a=\frac{v-u}{t}

where

u = 45 m/s is the initial velocity

v = 0 m/s is the final velocity

t = 10 s is the time taken

Substituting the numbers into the equation, we find

a=\frac{0 -45 m/s}{10 s}=-4.5 m/s^2

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For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro
serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf =  37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

               h = vot + 1/2gt²

             71.3 = 0t + 1/2(9.81)t²

             2(71.3) = 9,81t²

              t² = 2(71.3)/9.81

              t² = 14.53

              t = 3.81 s

b)

      vf = 0 + (9.81)(3.81)

      vf = 37.4 m/s

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An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan
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Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...

7 0
3 years ago
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