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When x = s = 11.2ft, the crate has a speed of 20 f

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Answer #1
Concepts and reason

The radius of curvature of the path can be calculated by using radius of curvature equation for path. The normal acceleration can be calculated if velocity and radius of curvature are known.

If the normal acceleration and tangential acceleration are known, the magnitude of the acceleration can be obtained using total acceleration equation. The direction the velocity can be calculated using slope equation of the path.

If the equation for the curve is given, without mentioning the value of radius of the curve then the radius of the curvature is calculated by radius of curvature equation.

If the body is moving in curvilinear motion, then the body has two acceleration components. One is normal acceleration and another one is tangential acceleration.

Fundamentals

Velocity is obtained by differentiation of displacement with respect to time. Velocity is a vector quantity that means it has direction. This direction is along the tangent to the path through which the object moves.

Here, s is the distance and time is t.

The magnitude of acceleration is derived by differentiation of velocity with respect to time. Acceleration is of two components. One is tangential acceleration and other is radial acceleration. Tangential acceleration is obtained by derivative. Radial acceleration is obtained by formula.

Consider the following formulas:

Here, radius of curvature is .

Radius of curvature is the reciprocal of curvature. Radius of curvature for curve is the distance from center of the curve top any point on curve. It is equal to radius in case of curves.

If the path is expressed as y= f(x)
, then the radius of curvature at any point on the path is calculated by using following expression:

The equation to calculate the magnitude of acceleration is,

Write the equation of the curve.

y =

Differentiate the above equation with respect to x.

记忆

Differentiate the above equation with respect to x.

dy_1
dx? 12

Let be the angle, the crate’s velocity is directed.

Calculate the direction of the crate’s velocity.

Tx=1121

Substitute for and 11.2t
for x.

$= tan (-1,2
12

$ = tan- (0.93333)
= 43.02°

Calculate the radius of curvature of the path as follows:

Ζιε L

Substitute for and for dy
.

(21/1)
zx[z(21/* ) +1]

Substitute 11.2t
for x.

[1+(11.2/12]]
(112)
[1.871]
(112)
= 30.7135f

Calculate the normal acceleration of the crate as follows:

Here, v is velocity of the crate.

Substitute 30.7135ft
for and 20 ft/s
for v.

(20ft/s)?
30.7135ft
= 13.0235ft/s

Writ the tangential acceleration of the crate.

a, = 5.6ft/s?

Calculate the magnitude of the crate acceleration as follows:

a=Ja +a?

Substitute 5.6ft/s?
for and 13.0235ft/s
for .

a=(5.6)’ +(13.0235)
a= 200.971 ft/s2
a = 14.176 ft/s?

Ans: Part A

The direction of the crate’s velocity is 43.029
.

Part B

The magnitude of the crate’s acceleration is 14.176ft/s
.

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