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Anettt [7]
3 years ago
6

Which word equation is used to calculate the acceleration of an object? A. Subtract the initial velocity from the final velocity

and multiply the result by the time. B. Subtract the initial velocity from the final velocity and divide the result by the time. C. Add the initial velocity and the final velocity and divide the result by the time. D. Add the initial velocity and the final velocity and multiply the result by the time.
Physics
2 answers:
saveliy_v [14]3 years ago
6 0

Answer:Subtract the initial velocity from the final velocity and divide the result by the time.

Explanation:

zlopas [31]3 years ago
3 0
Correct answer is: 
<span>B. Subtract the initial velocity from the final velocity and divide the result by the time

In fact, the formula to calculate the acceleration is
</span>a= \frac{v_f-v_i}{t}
<span>where vf is the final velocity, vi the initial velocity, and t the time.</span>
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A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the
jeka94

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

y = h + \frac{1}{2}at^2

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

v_x = 23.4 m/s

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

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(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

v_x = 23.4 m/s

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

v_y = u_y +at

where

u_y = 0 is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s

and the negative sign means it points downward.

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