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3 years ago

How to calculate the slope of position time graph?

Physics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

The slope of a position-time graph can be calculated as:

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

where

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

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

We can verify that the slope of this graph is actually equal to the velocity. In fact:

$\Delta y$ corresponds to the change in position, so it is the displacement, $\Delta s$

$\Delta x$ corresponds to the change in time $\Delta t$ , so the time interval

Therefore the slope of the graph is equal to

$m=\frac{\Delta s}{\Delta t}$

which corresponds to the definition of velocity.

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Read 2 more answers

1.a bag is dropped from a hovering helicopter. the bag has fallen for 2 s. what is the ball's velocity at the instant its hittin

omeli [17]

1. The bag's velocity immediately before hitting the ground.

Recall this kinematics equation:

Vf = Vi + aΔt

Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)

a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)

Δt = 2s

Plug in the values and solve for Vf:

Vf = 0 + 9.81×2

Vf = 19.62m/s

2. The height of the helicopter.

Recall this other kinematics equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.

Given values:

Vi = 0m/s (bag is dropped starting from rest)

a = 9.81m/s² (acceleration due to gravity of the earth)

Δt = 2s

Plug in the values and solve for d:

d = 0×2 + 0.5×9.81×2²

d = 19.62m

3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s

Reuse the equation from question 2:

d = ViΔt + 0.5aΔt²

Given values:

d = 19.6m (height of the helicopter obtained from question 2)

Vi = 2m/s

a = 9.81m/s² (acceleration due to earth's gravity)

Plug in the values and solve for Δt:

19.6 = 2Δt + 0.5×9.81Δt²

4.91Δt² + 2Δt - 19.6 = 0

Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):

Δt = 1.8s, Δt = −2.2s

The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.

Δt = 1.8s

Now we can use this new value of Δt to get the velocity before hitting the ground:

Vf = Vi + aΔt

Given values:

Vi = 2m/s

a = 9.81m/s²

Δt = 1.8s (result from previous question)

Plug in the values and solve for Vf:

Vf = 2 + 9.81×1.8

Vf = 19.66m/s

4 0

3 years ago

A ball is thrown from ground level so as to just clear wall 4m height at distance 4m from wall and falls 14 m from wall . Veloci

kolbaska11 [484]

Ok, this is a 2d kinematics problem, the falls 14 m part is confusing, I think it means in the x direction, but you don't need it anyway.

If we know it goes 4m into the air, we know d = 4m (height of wall), we also know the acceleration a=-9.8m/s^2 (because gravity) and that the vertical velocity when it just clears the wall will be 0 m/s, which we'll call our final velocity (Vf). Using Vf^2 = Vi^2 +2a*d, we can solve this for Vi and drop Vf because it's zero to get: Vi = sqrt(-2ad), plug in numbers (don't forget a is negative) and you get 8.85 m/s in the vertical direction. The x-direction velocity requires that we solve the y-direction for time, using Vf= Vi + at, we solve for t, getting t= -Vi/a, plug in numbers t= -8.85/-9.8 = 0.9 s. Now we can use the simple v = d/t (because x-direction has no acceleration (a=0)), and plug in the distance to the wall and the time it takes to get there v = (4/.9) = 4.444 m/s, this is the velocity in the x direction, we use Pythagoras' theorem to find the total velocity, Vtotal = sqrt(Vx^2 + Vy^2), so Vtotal = sqrt(8.85^2+4.444^2) = 9.9m/s. Yay physics!

8 0

3 years ago