   Chapter 4.9, Problem 66E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Show that for motion in a straight line with constant acceleration a, initial velocity v0, and initial displacement s0, the displacement after time t is s = 1 2 a t 2 + v 0 t + s 0

To determine

To show: The displacement of a straight line after time t is s=12at2+v0t+s0.

Explanation

Given:

Acceleration is a, initial velocity is v0, and initial displacement is s0.

Formula used:

The expression for acceleration function a(t).

a(t)=v(t) (1)

Here,

v(t) is first derivative of velocity function v(t).

The expression for velocity function v(t).

v(t)=s(t) (2)

Here,

s(t) is first derivative of displacement function s(t).

Antiderivative of t is 12t2 and 1 is t.

Calculation:

Initial velocity is v0, that is v(0)=v0 and the initial displacement is s0, that is s(0)=s0.

Given a(t)=a.

Substitute a for a(t) in equation (1),

v(t)=a

Integrate the equation v(t)=a with respect to t as follows.

v(t)=at+C (3)

Where C is an arbitrary constant.

Substitute 0 for t in equations (3),

v(0)=a(0)+C=C

Substitute v0 for v(0), C=v0

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