BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.8, Problem 2E

(a)

To determine

To find: The velocity at time t.

Expert Solution

Answer to Problem 2E

The velocity at time t is f'(t)=9(t29)(t2+9)2ft/sec_.

Explanation of Solution

Given:

The given equation is as below.

s=f(t)=9tt2+9 (1)

Calculation:

Calculate the velocity at time t.

Differentiate the equation (1) with respect to time by applying uv method.

f'(t)=(t2+9)(9)(9t×(2t))(t2+9)2=9t2+8118t2(t2+9)2=9t2+81(t2+9)2

f'(t)=9(t29)(t2+9)2 (2)

Therefore, the velocity at time t is f'(t)=9(t29)(t2+9)2ft/sec_.

(b)

To determine

To find: The velocity after 1 second.

Expert Solution

Answer to Problem 2E

The velocity after 1 second is v(1)=0.72ft/sec_.

Explanation of Solution

Calculate the velocity after 1 second.

Substitute 1 for t in the equation (2).

v(t)=f'(t)=9(t29)(t2+9)2v(1)=9(19)(1+9)2=0.72ft/sec

Therefore, the velocity after 1 second is v(1)=0.72ft/sec_.

(c)

To determine

To find: The time when particle at rest.

Expert Solution

Answer to Problem 2E

The time when particle at rest is t=3sec_.

Explanation of Solution

Calculate the time when particle at rest.

The velocity will be zero when the particle is at rest.

Substitute 0 for v(t) in the equation (2).

v(t)=9(t29)(t2+9)20=9(t29)(t2+9)29(t29)=0t29=0t=3sec

Therefore, the time when particle at rest is t=3sec_.

(d)

To determine

To find: The particle moving in the positive direction.

Expert Solution

Answer to Problem 2E

The particle always moves in positive direction when time is within the range 0t<3_.

Explanation of Solution

Calculate the time during which the particle will be moving in the positive direction.

If the particle moves in positive direction, the velocity at any time t will be greater than zero.

v(t)=9(t29)(t2+9)2>09(t29)(t2+9)2>09(t29)>0t29>0t<3

Therefore, the particle moving in the positive direction when 0t<3_.

(e)

To determine

To find: The total distance traveled during the first 6 seconds.

Expert Solution

Answer to Problem 2E

The total distance travelled during first 6 second is 1.8ft_.

Explanation of Solution

Calculate the total distance traveled during first 6 seconds.

Here, the particle moves in positive and negative direction, the total distance traveled should be calculated between the intervals of [0,3] and [3,6].

Substitute 3 and 0 for t between interval of [0,3] in the equation (1) and subtract them.

|f(3)f(0)|=|9(3)(3)2+99(0)(0)2+9|=|320|=32

Substitute 6 and 3 for t between intervals of [3,6] for t in the equation (1) and subtract them.

|f(6)f(3)|=|9(6)(6)2+99(3)(3)2+9|=|6532|=310

The total distance travelled is as given below.

f(6)=32+310=1.8ft

Therefore, the total distance travelled during first 6 second is f(6)=1.8ft_.

(f)

To determine

To find: The diagram to illustrate the motion of the particle.

Expert Solution

Explanation of Solution

Show the diagram to illustrate the motion of the particle as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.8, Problem 2E , additional homework tip  1

Figure 1 shows the movement of particle at different times.

(g)

To determine

To find: The acceleration at time t and after 1 second.

Expert Solution

Answer to Problem 2E

The acceleration at time t is f"(t)=18t(t227)(t3+9)3_ and after one second is 0.468ft/s2_.

Explanation of Solution

Calculate the acceleration at time t.

Differentiate the equation (2) with respect to t by applying uv method.

f'(t)=9(t29)(t2+9)2f"(t)=9(t2+9)2(2t)[(t29)(2(t2+9)×2t)][(t2+9)2]2=92t(t2+9)[(t2+9)2(t29)](t2+9)4

On further simplification,

f"(t)=18t[t2+92t2+18](t2+9)3=18t(t2+27)(t2+9)3=18t(t227)(t3+9)3

Therefore, the acceleration at time is f"(t)=18t(t227)(t3+9)3_.

Calculate the acceleration after 1 second.

Substitute 1 for t in the above equation.

f"(t)=18t(t227)(t3+9)3f"(1)=18(1)((1)227)(13+9)3=0.468fts2

Therefore, the acceleration after 1 second is f"(1)=0.468fts2_.

(h)

To determine

To find: The graph the position, velocity and acceleration function for 0t6.

Expert Solution

Explanation of Solution

Calculate the position of the particle with respect to time using the expression.

s(t)=9tt2+9

Substitute 0 for t in the above equation.

s(0)=9(0)02+9=0

Similarly, calculate the remaining values.

Calculate the value of t and s(t) as shown in the table (1).

ts(t)=9tt2+9
00
0.50.48649
10.9
1.51.2
21.38462
2.51.47541
31.5
3.51.48235
41.44
4.51.38462
51.32353
5.51.26115
61.2

Calculate the velocity using the formula.

v(t)=9(t29)(t2+9)2

Substitute 0 for t in the above equation.

v(0)=9(029)(02+9)2=1

Similarly, calculate the remaining values.

Calculate the value of t and v(t) as shown in the table (1).

tv(t)=9(t29)(t2+9)2
01
0.50.92038
10.72
1.50.48
20.26627
2.50.10642
30
3.5-0.0648
4-0.1008
4.5-0.1183
5-0.1246
5.5-0.1241
6-0.12

Calculate the acceleration using the formula.

a(t)=18t(t227)(t3+9)3

Substitute 0 for t in the above equation.

a(0)=18(0)(0227)(03+9)3=0

Similarly, calculate the remaining values.

Calculate the value of t and a(t) as shown in the table (1).

ta(t)=18t(t227)(t3+9)3
00.00
0.5-0.32
1-0.47
1.5-0.35
2-0.17
2.5-0.06
3-0.02
3.5-0.01
40.00
4.50.00
50.00
5.50.00
60.00

Draw the graph of the position, velocity and acceleration functions as shown in the Figure 2.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.8, Problem 2E , additional homework tip  2

(i)

To determine

To find: The time when particle speeding up and slowing down.

Expert Solution

Answer to Problem 2E

the acceleration is zero when time t is greater than 33sec and the acceleration is negative when time t is less than 33sec.

Explanation of Solution

Calculate the time when particle is speeding up and slowing down.

Substitute 0 for f"(t) in the equation f"(t)=18t(t227)(t3+9)3.

0=18t(t227)(t3+9)3t227=0t=27t=33sec

Substitute 1sec for t in the equation f"(t)=18t(t227)(t3+9)3.

f"(1)=18×(1)((1)227)((1)3+9)3=0.468<0

Substitute 4sec for t in the equation f"(t)=18t(t227)(t3+9)3.

f"(4)=18×(4)((4)227)((4)3+9)3=0

Therefore, the acceleration is zero when time t is greater than 33sec and the acceleration is negative when time t is less than 33sec.

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