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 3E

(a)

To determine

To find: The velocity at time t.

Expert Solution

Answer to Problem 3E

The velocity at time is f'(t)=π2cos(πt2)ft/sec_.

Explanation of Solution

Given:

The given equation is as below.

s=f(t)=sin(πt2) (1)

Calculation:

Calculate the velocity at time t.

Differentiate the equation (1) with respect to time.

v(t)=f'(t)=cos(πt2).π2

v(t)=f'(t)=π2cos(πt2)ftsec (2)

Therefore, the velocity at time t is f'(t)=π2cos(πt2)ft/sec_.

(b)

To determine

To find: The velocity after 1 second.

Expert Solution

Answer to Problem 3E

The velocity after 1 second is v(1)=0ft/sec_.

Explanation of Solution

Calculate the velocity after 1 second.

Substitute 1 for t in the equation (2).

v(t)=f'(t)=π2cos(πt2)v(1)=f'(1)=π2cos(π(1)2)v(1)=0ft/sec

Therefore, the velocity after 1 second is v(1)=0ft/sec_.

(c)

To determine

To find: The time when particle at rest.

Expert Solution

Answer to Problem 3E

The particle never is at rest when t=1+2n_.

Explanation of Solution

Calculate the time when particle will be at rest.

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

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

v(t)=π2cos(πt2)0=π2cos(πt2)cos(πt2)=0(πt2)=π2+nπt=(π2+nπ)2πt=1+2n

Here, n is the non-negative integer where t0.

(d)

To determine

To find: The particle moving in the positive direction.

Expert Solution

Answer to Problem 3E

The velocity of particle always moves in positive direction when 0<t<1_,3<t<5_,7<t<9_.

Explanation of Solution

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

The particle will move in positive direction when v(t)>0, and the equation from part (c), the velocity changes its sign at every positive odd integer. Therefore, the intervals are 0<t<1,3<t<5,7<t<9 and so on.

(e)

To determine

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

Expert Solution

Answer to Problem 3E

The total distance travelled during first 6 second is f(6)=6ft_.

Explanation of Solution

Calculate the total distance traveled during first 6 seconds.

Here, the velocity changes from 1, 3 and 5 which appears in the interval of [0,6].

Substitute 1 and 0 for t in the equation (1).

|f(1)f(0)|=|sinπ(1)2sinπ(0)2|=|10|=1|f(3)f(1)|=|sinπ(3)2sinπ(1)2|=|11|=2|f(5)f(3)|=|sinπ(5)2sinπ(3)2|=|1(1)|=2|f(6)f(5)|=|sinπ(6)2sinπ(5)2|=|01|=1

The total distance travelled is as below.

f(t)=1+2+2+1=6ft

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

(f)

To determine

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

Expert Solution

Answer to Problem 3E

The diagram to illustrate the motion of particle is shown in the figure (1).

Explanation of Solution

Calculate the distance s using the expression.

s=sinπt2

Substitute 0 for t in the above equation.

s=sinπt2s=sinπ(0)2s=0

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

ts
00
11
3-1
51
60

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 3E , additional homework tip  1

(g)

To determine

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

Expert Solution

Answer to Problem 3E

The acceleration at time is π24sinπt2ft/s2_ and after one second is π24ft/s2_.

Explanation of Solution

Calculate the acceleration at any time t.

Differentiate the equation (2) with respect to t.

a(t)=π2(sinπt2×π2)=π24sinπt2ft/s2

Therefore, the acceleration at any time is π24sinπt2ft/s2_.

Calculate the acceleration after 1 second.

Substitute 1 for t in the above equation a(t)=π2(sinπt2×π2).

a(1)=π24sinπ(1)2ft/s2=π24ft/s2

Therefore, the acceleration after 1 second is π24ft/s2_.

(h)

To determine

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

Expert Solution

Answer to Problem 3E

The position, velocity, and acceleration functions are plotted for time limits 0t6 in figures (2), (3), and (4).

Explanation of Solution

Calculate the position with respect to time using the formula.

f(t)=sinπt2

Substitute 0 for t in the above equation.

f(t)=sinπt2f(0)=sinπ(0)2=0

Similarly, calculate the remaining values.

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

tf(t)=sinπt2
0.000.00
0.500.71
1.001.00
1.500.71
2.000.00
2.50-0.71
3.00-1.00
3.50-0.71
4.000.00
4.500.71
5.001.00
5.500.71
6.000.00

Calculate the velocity using the expression.

v(t)=π2cosπt2

Substitute 0 for t in the above equation.

v(0)=π2cosπ(0)2=1.57

Similarly, calculate the remaining values.

Calculate the value of t and v(t) as shown in the table (2).

tv(t)=π2cosπt2
0.001.57
0.501.11
1.000.00
1.50-1.11
2.00-1.57
2.50-1.11
3.000.00
3.501.11
4.001.57
4.501.11
5.000.00
5.50-1.11
6.00-1.57

Calculate the acceleration using the formula.

a(t)=π24sinπt2

Substitute 0 for t in the above equation.

a(t)=π24sinπt2a(0)=π24sinπ(0)2=0

Similarly, calculate the remaining values.

Calculate the value of t and a(t) as shown in the table (3).

ta(t)=π24sinπt2
0.000.00
0.50-1.74
1.00-2.47
1.50-1.74
2.000.00
2.501.74
3.002.47
3.501.74
4.000.00
4.50-1.74
5.00-2.47
5.50-1.74
6.000.00

Draw the position as a function of time curve as shown in the Figure (2).

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.8, Problem 3E , additional homework tip  2

Draw the speed as a function of time curve as shown in the figure (3).

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.8, Problem 3E , additional homework tip  3

Draw the acceleration as a function of time curve as shown in the figure (4).

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.8, Problem 3E , additional homework tip  4

(i)

To determine

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

Expert Solution

Answer to Problem 3E

The acceleration is negative when the value of time is 0<t<2 seconds and this will be slowing down and the acceleration is positive when the value of time is 2<t<4sec and this will be speeding up. This cycle will repeat in the next 4 seconds, and go on.

Explanation of Solution

Calculate the time when particle speeding up and slowing down.

Substitute 0 for f"(t) in the equation f"(t)=π24sinπt2.

0=π24sinπt2sinπt2=0t=0,2,4,...2nsec

Here, variable n is given as natural numbers starting from 0.

Therefore, the acceleration is negative when the value of time is 0<t<2 seconds and this will be slowing down and the acceleration is positive when the value of time is 2<t<4sec and this will be speeding up. This cycle will repeat in the next 4 seconds, and go on.

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