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galina1969 [7]
2 years ago
11

Describe how different types of motion are represented by distance-time graphs and velocity-time graphs.

Physics
1 answer:
eduard2 years ago
4 0

Answer:

non-accelerated movement

velocity versus time  a horizontal straight line.

distance versus time  gives a horizontal straight line.

accelerated motion

graph of velocity versus time s an inclined line and the slope

graph of distance versus time is a parabola of the form

Explanation:

In kinematics there are two types of steely and non-accelerated movements

In a  the velocity of the body is constant therefore a speed hook against time gives a horizontal straight line.

A graph of distance versus time is a straight line whose slope is the velocity of the body

          x = v t

In an accelerated motion the velocity changes linearly with time, so a graph of velocity versus time is an inclined line and the slope is the value of the acceleration of the body

         v = v₀ + a t

A graph of distance versus time is a parabola of the form

         x =v₀ t + ½ a t²

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The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

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The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

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