# The velocity at time t .

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 3.8, Problem 3E

(a)

To determine

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

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

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

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

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

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).

 t s 0 0 1 1 3 -1 5 1 6 0

Show the diagram to illustrate the motion of the particle as shown below in figure (1).

(g)

To determine

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

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).

 t f(t)=sinπt2 0.00 0.00 0.50 0.71 1.00 1.00 1.50 0.71 2.00 0.00 2.50 -0.71 3.00 -1.00 3.50 -0.71 4.00 0.00 4.50 0.71 5.00 1.00 5.50 0.71 6.00 0.00

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).

 t v(t)=π2cosπt2 0.00 1.57 0.50 1.11 1.00 0.00 1.50 -1.11 2.00 -1.57 2.50 -1.11 3.00 0.00 3.50 1.11 4.00 1.57 4.50 1.11 5.00 0.00 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).

 t a(t)=−π24sinπt2 0.00 0.00 0.50 -1.74 1.00 -2.47 1.50 -1.74 2.00 0.00 2.50 1.74 3.00 2.47 3.50 1.74 4.00 0.00 4.50 -1.74 5.00 -2.47 5.50 -1.74 6.00 0.00

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

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

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

(i)

To determine

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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