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Kipish [7]
3 years ago
9

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

t the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Engineering
1 answer:
irina [24]3 years ago
8 0

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car  to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both  cars which is equal from instant point of abreast.

So; we can say :

D_{pursuit} =D_{police}

By using the second equation of motion to find the distance S;

S= ut + \dfrac{1}{2}at^2

D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)

D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)

D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)

D_{police} = ut _P + \dfrac{1}{2}at_p^2

where ;

u  = 0

D_{police} =  \dfrac{1}{2}at_p^2

D_{police} =  \dfrac{1}{2}*(1.96)*(t+12)^2

D_{police} = 0.98*(t+12)^2

D_{police} = 0.98*(t^2 + 144 + 24t)

D_{police} = 0.98t^2 + 141.12 + 23.52t

Recall that:

D_{pursuit} =D_{police}

(187.8+15.65 \ t-1.525 t^2)=  0.98t^2 + 141.12 + 23.52t

(187.8 - 141.12)  + (15.65 \ t  -  23.52t)  -( 1.525 t^2    - 0.98t^2)  =   0

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR  t = 3.02

Since we can only take consideration of the value with a  positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time  required for the police car  to over take the automobile = 12 s + 3.02 s

Total time  required for the police car  to over take the automobile = 15.02 sec

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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

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True Stress

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     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

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Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

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From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

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       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

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