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Solid mechanics with a positive reference

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1. A 2-lb ball is thrown upward at 30ft/s from an origin 5-ft above the earth's surface. Using for your positive reference a vertical Y axis with its origin at 5ft, DETERMINE FOR THE BALL:

a) The maximum height in ft reached
b) The elapsed time of flight in seconds until its return to the origin
c) Its acceleration in ft/s2 at the highest point

2. The velocity of a 0.1kg particle along a horizontal X axis is given by:
v(t)=(8t2 -20)m/s, and at t=20s, v(20)=60m. DETERMINE FOR t=8s:

a) The position of the particle in ft
b) The velocity of the particle in ft/s
c) The acceleration of the particle in ft/s2

3. The acceleration of an object in rectilinear x-direction motion is given by:
a(v)=0.4-0.0002v2 ft/s. If v=0 when x=0. DETERMINE FOR THE OBJECT:

e) Its velocity in ft/s when x=1000ft
f) Its acceleration in ft/s2 at x=1000ft
g) Its maximum velocity in ft/s

https://brainmass.com/physics/velocity/solid-mechanics-positive-reference-161165

SOLUTION This solution is FREE courtesy of BrainMass!

The velocity , where s is the displacement of the object,
And the acceleration

On the other hand, we have
The velocity is
The displacement

1. A 2-lb ball is thrown upward at 30ft/s from an origin 5-ft above the earth's surface. Using for your positive reference a vertical Y axis with its origin at 5ft, DETERMINE FOR THE BALL:

a) The maximum height in ft reached
b) The elapsed time of flight in seconds until its return to the origin
c) Its acceleration in ft/s2 at the highest point

The ball is subject to the gravitational acceleration g all the time.
When the positive direction is upward, the acceleration a of the ball is
a = - g = - 32 ft/s2.

So the velocity is
, where c1 is the integration constant, which is determined by the initial velcotiy.
, so
The velocity function is then
ft/s.

The height is

When t = 0, h = 0, so
The height function is

(a) the maximum height:
First find the critical point
s
Now according to the second derivative test, , so at , the height function h(t) has a maximum.
The maximum height is
ft.
The maximum height is 14.0625 ft from the defined origin.
It is 14.0625 + 5 = 19.0625 ft high from the earth's surface.

(b) when the ball returns back to the origin, the height h(t) = 0

Apparently, it takes 1.875 seconds for the ball to return back to the origin.

(c) the ball has a constant acceleration of -32 ft/s2. so the acceleration is -32 ft/s2 at the highest point.

(d) when t = 1.875 s, the ball returns back to the origin, so
ft/s.
The velocity is -30 ft/s when returning back to the origin.

2. The velocity of a 0.1kg particle along a horizontal X axis is given by:
v(t)=(8t2 -20)m/s, and at t=20s, v(20)=60m. DETERMINE FOR t=8s:
Apparently, when t = 20 s, the velocity v(20) = 3180 m/s. So I think the red-highlighted parts should be t = 20 s, d(20) = 60 m, where d is the position of the particle.
a) The position of the particle in ft

When t = 20, d(20) = 60, so

The position is
meter
So when t = 8 s,
meter.
The position of the particle is -19668 meters at t = 8.

b) The velocity of the particle in ft/s
When t = 8 s,
m/s
c) The acceleration of the particle in ft/s2
The acceleration is
, unit: m/s2
When t = 8 s
m/s2.

All units are in meters, you can convert to ft using
1 m =
For example, the position of the particle is then
The velocity is
The acceleration is

3. The acceleration of an object in rectilinear x-direction motion is given by:
a(v)=0.4-0.0002v2 ft/s. If v=0 when x=0. DETERMINE FOR THE OBJECT:

e) Its velocity in ft/s when x=1000ft
f) Its acceleration in ft/s2 at x=1000ft
g) Its maximum velocity in ft/s

The acceleration
So we have

Integrate on both sides

When x = 0, v = 0, so

Take the exponential on both sides, we have

(a) when x = 1000 ft,

Then the velocity is
or -25.68 ft/s

(b) the acceleration is

When x = 1000 ft,

(c) The velocity function is

We know that the exponential function is always positive, so when it is at its minimum, is at is maximum, that is, is at its maximum.

So the maximum of v2 is

That maximum speed is 44.72 ft/s.

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