VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 11.3, Problem 11.61P

A particle moves in a straight line with a constant acceleration of −4 ft/s2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of +4 ft/s2 for the next 4 s. Knowing that the particle starts from the origin and that its velocity is −8 ft/s during the zero acceleration time interval, (a) construct the vt and xt curves for 0 ≤ t ≤ 14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 14 s.

Fig. P11.61 and P11.62

Chapter 11.3, Problem 11.61P, A particle moves in a straight line with a constant acceleration of 4 ft/s2 for 6 s, zero

(a)

Expert Solution
Check Mark
To determine

Construct the vt and xt curves for 0t14sec.

Explanation of Solution

Given information:

The constant acceleration (a1) of particle at 6 sec is 4ft/s2.

The acceleration is zero from 6sec to 10 sec.

From 10 sec to 14 sec the acceleration (a2) of the particle is 4ft/s2.

The velocity (v6) of the particle at 6 sec is 8ft/s.

Calculation:

Show a-t curve of particle that moves in a straight line as in Figure (1).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 11.3, Problem 11.61P , additional homework tip  1

Calculate the area (A1) using the relation:

A1=t1a1

Substitute 6 sec for t1 and 4ft/s2 for a1.

A1=(6)×(4)=24ft/s

Calculate the area (A2) using the relation:

A2=t3a2

Substitute 4 sec for t3 and 4ft/s2 for a2.

A2=(4)×(4)=16ft/s

Calculate the velocity (v0) at the origin using the relation:

v0=v6A1

Substitute 8ft/s for v6 and 24ft/s for A1.

v0=8(24)=16ft/s

Calculate the velocity (v10) at 10 sec using the relation;

v10=v6

Substitute 8ft/s for v6.

v10=8ft/s

Calculate the velocity (v14) at 14 sec using the relation:

v14=v10+A2

Substitute 8ft/s for v10 and 16ft/s for A2.

v14=8+16=8ft/s

Tabulated the acceleration (a), velocity (v) corresponding to time (t) in Table (1) :

t(s)a(ft/s2)v(ft/s)
0-416
60-8
100-8
1448

Plot the v-t curve of particle that moves in a straight line with areas as in Figure (2).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 11.3, Problem 11.61P , additional homework tip  2

Calculate the area (A3) using the relation:

A3=12v0t4

Here, t4 is time between 0 sec to 4 sec.

Substitute 4 sec for t4 and 16ft/s for v0.

A3=12(16)(4)=32ft

Calculate the area (A4) using the relation:

A4=12v6t5

Here, t5 is time between 4 sec to 6 sec.

Substitute 2 sec for t5 and 8ft/s for v6.

A4=12(8)(2)=8ft

Calculate the area (A5) using the relation:

A5=t6v10

Here, t6 is time between 6 sec to 10 sec

Substitute 4 sec for t6 and 8ft/s for v10.

A5=(4)×(8)=32ft

Calculate the area (A6) using the relation:

A6=12v10t7

Here, t7 is time between 10 sec to 12 sec.

Substitute 2 sec for t7 and 8ft/s for v10.

A6=12(8)(2)=8ft

Calculate the area (A7) using the relation:

A7=12v14t8

Here, t8 is time between 12 sec to 14 sec.

Substitute 4 sec for t8 and 8ft/s for v14.

A7=12×8×2=8ft

Calculate the position (x0) of the particle at 0 sec using the relation:

x0=0ft

Calculate the position (x4) of the particle at 4 sec using the relation:

x4=x0+A3

Substitute 0 for x0 and 32ft for A3.

x4=0+32=32ft

Calculate the position (x6) of the particle at 6 sec using the relation:

x6=x4+A4

Substitute 32ft for x4 and 8ft for A4.

x6=328=24ft

Calculate the position (x10) of the particle at 10 sec using the relation:

x10=x6+A5

Substitute 24ft for x6 and 32ft for A5.

x10=2432=8ft

Calculate the position (x12) of the particle at 12 sec using the relation:

x12=x10+A6

Substitute 8ft for x10 and 8ft for A6.

x12=88=16ft

Calculate the position (x14) of the particle at 14 sec using the relation:

x14=x12+A7

Substitute 16ft for x12 and 8ft for A7.

x14=16+8=8ft

Tabulated the position (x) corresponding to time (t) in Table 2:

t (sec)x (ft)
00
432
624
10-8
12-16
14-8

Plot x-t curve of particle that moves in a straight line with areas as in Figure 3.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 11.3, Problem 11.61P , additional homework tip  3

(b)

Expert Solution
Check Mark
To determine

The position, velocity of the particle and the total distance (d) traveled when time (t) 14 sec.

Answer to Problem 11.61P

The total distance (d) traveled when time (t) 14 sec is 88ft_.

Explanation of Solution

Given information:

The constant acceleration (a1) of particle at 6 sec is 4ft/s2.

The acceleration is zero from 6sec to 10 sec.

From 10 sec to 14 sec the acceleration (a2) of the particle is 4ft/s2.

The velocity (v6) of the particle at 6 sec is 8ft/s.

Calculation:

Calculate the area (A1) using the relation:

A1=t1a1

Substitute 6 sec for t1 and 4ft/s2 for a1

A1=(6)×(4)=24ft/s

Calculate the area (A2) using the relation:

A2=t3a2

Substitute 4 sec for t3 and 4ft/s2 for a1

A2=(4)×(4)=16ft/s

Calculate the velocity (v0) at the origin using the relation:

v0=v6A1

Substitute 8ft/s for v6 and 24ft/s for A1.

v0=8(24)=16ft/s

Calculate the velocity (v10) at 10 sec using the relation;

v10=v6

Substitute 8ft/s for v6.

v10=8ft/s

Calculate the velocity (v14) at 14 sec using the relation:

v14=v10+A2

Substitute 8ft/s for v10 and 16ft/s for A2.

v14=8+16=8ft/s

Calculate the area (A3) using the relation:

A3=12v0t4

Here, t4 is 0 sec to 4 sec.

Substitute 4 sec for t4 and 16ft/s for v0.

A3=12(16)(4)=32ft

Calculate the area (A4) using the relation:

A4=12v6t5

Here, t5 is 4 sec to 6 sec.

Substitute 2 sec for t5 and 8ft/s for v6.

A4=12(8)(2)=8ft

Calculate the area (A5) using the relation:

A5=t6v10

Here, t6 is 6 sec to 10 sec

Substitute 4 sec for t6 and 8ft/s for v10.

A5=(4)×(8)=32ft

Calculate the area (A6) using the relation:

A6=12v10t7

Here, t7 is 10 sec to 12 sec.

Substitute 2 sec for t7 and 8ft/s for v10.

A6=12(8)(2)=8ft

Calculate the area (A7) using the relation:

A7=12v14t8

Here, t8 is 12 sec to 14 sec.

Substitute 4 sec for t8 and 8ft/s for v14.

A7=12×8×2=8ft

Calculate the position (x0) of the particle in 0 sec using the relation:

x0=0ft

Calculate the position (x4) of the particle in 4 sec using the relation:

x4=x0+A3

Substitute 0 for x0 and 32ft for A3. 0

x4=0+32=32ft

Calculate the position (x6) of the particle in 6 sec using the relation:

x6=x4+A4

Substitute 32ft for x4 and 8ft for A4.

x6=328=24ft

Calculate the position (x10) of the particle in 10 sec using the relation:

x10=x6+A5

Substitute 24ft for x6 and 32ft for A5.

x10=2432=8ft

Calculate the position (x12) of the particle in 12 sec using the relation:

x12=x10+A6

Substitute 8ft for x10 and 8ft for A6.

x12=88=16ft

Calculate the position (x14) of the particle in 14 sec using the relation:

x14=x12+A7

Substitute 16ft for x12 and 8ft for A7.

x14=16+8=8ft

Calculate the distance (d1) traveled when time is 0t4sec using the relation:

d1=|x4x0|

Substitute 0 for x0 and 32ft for x4.

d1=|320|=32ft

Calculate the distance (d2) traveled when time is 4sect12sec using the relation:

d2=|x12x4|

Substitute 32ft for x4 and 16ft for x12.

d2=|1632|=|48|=48ft

Calculate the distance (d3) traveled when time is 12sect14sec using the relation:

d3=|x14x12|

Substitute 16ft for x12 and 8ft for x14.

d3=|8(16)|=8ft

Calculate the total distance (d) traveled when time (t) is 14 sec

d=d1+d2+d3

Substitute 32ft for d1, 48ft for d2 and 8ft for d3.

d=32+48+8=88ft

Therefore, the total distance (d) traveled when time (t) 14 sec is 88ft_.

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Chapter 11 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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