# The velocity and acceleration 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 3.3, Problem 35E

(a)

To determine

## To find: The velocity and acceleration at time t.

Expert Solution

The velocity of smooth level surface is 8cost_.

The acceleration of smooth level surface is 8sint_.

### Explanation of Solution

Given:

The equation of motion is x(t)=8sint, where t is in seconds and x is in centimeters.

Derivative rule:

Constant Multiple Rule:

If c is constant and f(x). is differentiable function, then

ddx[cf(x)]=cddx[f(x)] (1)

Recall:

If x(t)x is a displacement of a particle and the time t is in seconds, then the velocity of the particle is v(t)=dxdt.

If v(t) is a velocity of the particle and the time t is in seconds, then the acceleration of the particle is a(t)=dvdt.

Calculation:

Obtain the velocity at time t.

v(t)=ddt(x(t))=ddt(8sint)

Apply the Constant Multiple Rule as shown in equation (1),

v(t)=8ddx(sint)=8cost

Thus, the velocity of  x(t)=8sint is v(t)=8cost.

Obtain the acceleration at time t.

a(t)=ddt(v(t))=ddt(8cost)

Apply the Constant Multiple Rule as shown in equation (1),

a(t)=8ddx(cost)=8(sint)=8sint

Therefore, the acceleration of v(t)=8cost is a(t)=8sint.

(b)

To determine

### To find: The position, velocity and acceleration at t=2π3 and obtain its direction.

Expert Solution

The position, velocity and acceleration at t=2π3 are 43, 4 and 43, respectively.

The direction of the particle is moving to the left (negative direction).

### Explanation of Solution

Given:

The equation of motion is x(t)=8sint, where t is in seconds and x is in centimeters.

Calculation:

The position x(t)=8sint at t=2π3 is computed as follows,

x(2π3)=8sin2π3=8(32)      (Q sin2π3=32 )=43

Thus, the position x(t)=8sint at t=2π3 is x(2π3)=43.

From part (a), the velocity of  x(t) is v(t)=8cost.

The velocity v(t)=8cost at t=2π3 is computed as follows,

v(2π3)=8cos(2π3)=8(12)      (Q   cos2π3=12)=4

Thus, the velocity v(t)=8cost at t=2π3 is v(2π2)=4.

From part (a), the acceleration of v(t)=8cost is a(t)=8sint.

The acceleration a(t)=8sint at t=2π3 is computed as follows,

a(2π3)=8sin2π3=8(32)     (Q sin2π3=32 )=43

Thus, the acceleration a(t)=8sint at t=2π3 is a(2π3)=43.

Here, the velocity of the particle at t=2π3 is less than zero. That is, v(2π3)<0.

This implies that, the velocity of the particle at t=2π3 is negative.

Therefore,the direction of the particle is moving to the left (negative direction).

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