Question

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.

Answer 1

Answer:

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