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

a) The jogger has run approximately 7.978 miles after 1 hour. b) The jogger's acceleration at t = 0 is approximately 0.41 ft/s^2. c) The equation does not provide a direct relationship between distance and time, so we cannot determine the time required to run 6 miles.

(a) To find the distance the jogger has run when t = 1 h, we need to substitute t = 1 into the given equation. Plugging in t = 1, we have: v = 7.5(1 + 2(0.04(1)))^0.3. Simplifying this expression, we get v ≈ 7.5(1.08)^0.3. Evaluating this expression, we find v ≈ 7.978 mi/h. To find the distance, we multiply the speed by the time: d = v × t = 7.978 × 1 = 7.978 mi.

(b) The acceleration can be found by taking the derivative of the velocity equation with respect to time. Differentiating the given equation, we have dv/dt = 7.5(1 + 2(0.04x))^(-0.7) × 2(0.04). Substituting t = 0 into this expression, we get: a = 7.5(1 + 2(0.04(0)))^(-0.7) × 2(0.04). Simplifying this expression, we find a ≈ 0.28 mi/h^2. To convert this to ft/s^2, we multiply by 5280 ft/mi and divide by 3600 s/h: a = 0.28 × 5280 / 3600 ≈ 0.41 ft/s^2.

(c) To find the time required for the jogger to run 6 mi, we need to solve the equation v = 7.5(1 + 2(0.04x))^0.3 for t. Plugging in d = 6 and v = 7.5(1 + 2(0.04x))^0.3, we can solve for t. However, the given equation does not provide a direct relationship between distance and time, so we cannot solve it.

Answer 2

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