Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5 (1 2 0.04x) 0.3, where v and x are expressed in mi/h and miles, respectively. Knowing that x 5 0 at t 5 0, determine (a) the distance the jogger has run when t 5 1 h, (b) the jogger's acceleration in ft/s2 at t 5 0, (c) the time required for the jogger to run 6 mi.

solution:

to find the speed of a jogger use the following relation:

V

=

d

x

/d

t

=

7.5

*m

i

/

h

r

... (

1

)

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

*=

∫

d

t

+

C

or

-

4.7619

=

t

+

C

Here C = constant of integration.

x

=

0 at t

=

0

, we get: C

=

-

4.7619

now we have the relation to find the position and time for the jogger as:

-

4.7619 =

t

-

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= -

4.7619

=

1

-

4.7619

= -

3.7619

or x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in m

i

l

/

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

v

=

7.5*

Now differentiating above equation w. r. t time we get:

a

=

d

v/

d

t

=

-

0.675

/

At

t

=

0

the joggers acceleration is:

a

=

-

0.675

m

i

l

/

=

-

4.34

*

f

t

/

(c) required time for the jogger to run 6 miles is obtained by setting

x

=

6 in equation (2). We get:

-

4.7619

(

1

-

(

0.04

*

6)

) ^

7

/

10=

t

-

4.7619

or

t

=

0.832

h

r

s