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 4.8, Problem 46E
To determine

whether the balls ever pass each other.

Expert Solution

Answer to Problem 46E

The balls pass each other at the time of t=5seconds_ .

Explanation of Solution

Given data:

Consider the edge of cliff is at 432 ft above the ground.

Speed of first ball is 48fts and speed of second ball is 24fts .

Formula Used:

Write the expression for acceleration function a(t) .

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

Here,

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

Write 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 of first ball is 48fts , that is v1(0)=48fts and second ball is thrown after a second with velocity of 24fts , that is at v2(1)=24fts .

The two balls are thrown from same edge and edge of cliff is at 432 ft. Hence,

s1(0)=s2(0)=432

The motion of stone is close to ground, so the motion is considered as gravitational constant (g) , which is 32fts2 .

Write the expression for acceleration function of first ball (a(t)) .

a1(t)=32

Substitute v1(t) for v(t) and 32 for a(t) in equation (1),

v1(t)=32

Antiderivate the expression with respect to t,

v1(t)=32t+C (3)

Here,

C is arbitrary constant.

Substitute 0 for t in equations (3),

v1(0)=32(0)+C=C

Substitute 48 for v1(0) ,

C=48

Substitute 48 for C in equation (3),

v1(t)=32t+48 (4)

Substitute 32t+48 for v(t) in equation (2),

s1(t)=32t+48

Antiderivate the expression with respect to t,

s1(t)=32(12t2)+48t+D

s1(t)=16t2+48t+D (5)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (5),

s1(0)=16(0)2+48(0)+D=0+D=D

Substitute 432 for s1(0) ,

D=432

Substitute 432 for D in equation (5),

s1(t)=16t2+48t+432

Write the expression for acceleration function of second ball (a2(t)) .

a2(t)=32

Substitute v2(t) for v(t) and 32 for a(t) in equation (1),

v2(t)=32

Antiderivate the expression with respect to t,

v2(t)=32t+C (6)

Substitute 1 for t in equations (6),

v2(1)=32(1)+C=32+C

Substitute 24 for v2(1) ,

24=32+CC=24+32C=56

Substitute 56 for C in equation (6),

v1(t)=32t+48 (7)

Substitute 32t+48 for v(t) in equation (2),

s2(t)=32t+56

Antiderivate the expression with respect to t,

s2(t)=32(12t2)+56t+D

s2(t)=16t2+56t+D (8)

Substitute 1 for t in equation (8),

s2(1)=16(1)2+56(1)+D=16+56+D=40+D

Substitute 432 for s2(1) ,

432=40+DD=43240D=392

Substitute 392 for D in equation (8),

s2(t)=16t2+56t+392

The balls pass each other when their positions are equal.

s1(t)=s2(t)

Substitute 16t2+48t+432 for s1(t) and 16t2+56t+392 for s2(t) ,

16t2+48t+432=16t2+56t+39216t2+48t+432+16t256t392=08t+40=08t=40

t=408secondst=5seconds

Thus, the balls pass each other at the time of t=5seconds_ .

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