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ipn [44]
3 years ago
8

HELP PLEASE

Physics
2 answers:
Leto [7]3 years ago
8 0

Answer:

2.5m/s^2

Explanation:

We have the first equation of motion here:

V(f) = V(i) + at (equation 1)

Now,

V(f) = Final velocity = 25 m/s

V(i) = Initial velocity = 0 m/s (Since car starts moving from rest, so intial velocity has to be considered as 0 m/s)

a = Acceleration = ? (It has to be determined)

t= Time = 10 seconds

Now by putting values in equation 1

25 = 0 + (a)*(10)

25 = 10*a

a = 25/10

a = 2.5 m/s^2

Additionally, remember that we have three equations of motion

First equation is V(f) = V(i) + at

Second is S = V(i)t + (1/2)a(t^2)

Third is 2as = [V(f)]^2 - [V(i)]^2

Where S = Distance

            t = Time

            a= Acceleration

            V(f) = Final velocity

            V(i) = Initial velocity

We always decide to use any of these equations according to the data given, like you can see that I used first equation because it is independent of Distance, since in data, there was no distance given, so i used first equation of motion.

4vir4ik [10]3 years ago
7 0

Recall this kinematic equation:

a = \frac{Vi+Vf}{Δt}

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s


Substitute the terms in the equation with the given values and solve for a:

a = \frac{0+25}{10}

<h3>a = 2.5 m/s²</h3>
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yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

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The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

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Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

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Solutions

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     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0
3 years ago
( Can someone help? )
Murrr4er [49]

Answer:

Answer would be 0.33

Explanation:

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8 0
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Answer:

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2gd = u² - v²

d = \frac{u^{2}-v^{2}  }{2g}\\\\d = \frac{7.5^{2}-2.2^{2}  }{2*9.8}\\\\d = 2.62 \ m

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