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3 years ago

How can you determine an object's average velocity from a position time graph

Physics

1 answer:

viktelen [127]3 years ago

4 0

The average velocity can be calculated from the slope of the graph

Explanation:

In a position-time graph, the position is represented on the y-axis while the time is represented on the x-axis.

The slope of the graph is given by:

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the change in the y-variable

$\Delta x$ is the change in the x-variable

However, we have:

$\Delta y$ corresponds to the change in position, therefore it is the displacement
$\Delta x$ corresponds to the change in time

This means that the slope can be rewritten as

$m=\frac{displacement}{time}$

However, this ratio corresponds to the definition of average velocity:

$velocity = \frac{displacement}{time}$

Therefore, the slope of a position-time graph gives the average velocity.

Learn more about velocity:

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lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of

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The intensity of laser 2 is 4 times of the intensity of laser 1.

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An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

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Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

τ = 26.58 Nm

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

τ = 18.79 Nm

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