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Kisachek [45]
4 years ago
15

What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m

Physics
1 answer:
Pepsi [2]4 years ago
5 0

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of
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1. 27.3 m/s

The velocity of the gazelle at any time is given by:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

a=4.2 m/s^2

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

v=0+(4.2)(6.5)=27.3 m/s

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m

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d=\frac{1}{2}at^2

And substituting d = 30 m, we find the time:

t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(30)}{4.2}}=3.8 s

3. 10.6 s

In this case, the  distance the gazelle must cover is 200 m.

We know that in the first 6.5 s, the gazelle covers a distance of 88.7 m.

In the second part of the motion, the gazelle continues at its top speed, which is:

v = 27.3 m/s

The gazelle still have to cover a distance of

d' = 200-88.7 =111.3 m

Therefore, the time taken to cover this distance is

t'=\frac{d'}{v}=\frac{111.3}{27.3}=4.1 s

So, the total time the gazelle needs to cover 200 m is

t = 6.5 + 4.1 = 10.6 s

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Heat transfer through Radiation is responsible.

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The fuel will be for all the city is 2,000 kilograms
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Calculate the net force experienced by Mr. Ferrari if he has a mass of 100 kg and an acceleration of 3.5 m/s/s.
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