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Andrew [12]
3 years ago
7

It took Eric 11 hours to drive to a football game. On the way home, he was able to increase his average speed by 12mph and make

the return drive in only 9 hours. Find his average speed on the return drive.
Physics
1 answer:
yarga [219]3 years ago
6 0

Answer:

54mph

Explanation:

Speed is defined as the change in distance of a body with respect to time.

Let x be the speed of Eric on his way to the football game, if he used 11hours to the football game, his distance will be the product of his speed and time taken i.e

Distance = speed × time

Distance= x × 11

Distance = 11x... (1)

During his return journey, his speed increases by 12mph with a return drive in only 9hours i.e speed = x+12 and time is 9hours

Since distance = speed × time (return Journey)

Distance = (x+12)×9

Distance = 9x+84...(2)

Note that the distance to and fro the journey is the same as such we will equate equations 1 and 2 to get his initial speed "x" to have;

11x = 9x+84

11x-9x = 84

2x = 84

x = 42mph

This means that his initial speed on his way to the football game is 42mph. Since his speed increases by 12mph during his return drive, his return speed will be 42mph + 12mph which gives 54mph.

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Current and voltage on each resistor:

I_1 = 3.98 A, V_1 = 3.98 V

I_2=0.015 A, V_2 = 0.075 V

I_3 = 0.4 A, V_3 = 0.4 V

I_4 = 0.385 A, V_4 = 0.77 V

I_5 = 0.585 A, V_5 = 1.17 V

I_6 = 3.01 A, V_6 = 6.02 V

I_7 = 0.97 A, V_7 = 4.85 V

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega

This resistor is in series with resistor 4, so:

R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega

This resistor is in parallel with resistor 5, therefore:

R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega

This resistor is in series with resistor 7, so:

R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega

This resistor is in parallel with resistor 6, so:

R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega

Finally, this combination is in series with resistor 1:

R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A

And we can also find the potential difference across resistor 1:

V_1=I_1 R_1=(3.98)(1.0)=3.98 V

This means that the voltage across resistor 6 is

V_6=V-V_1=10-3.98=6.02 V

And so, the current on resistor 6 is

I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

I_7=I-I_6=3.98-3.01=0.97 A

And so the voltage across resistor 7 is

V_7=I_7 R_7=(0.97)(5.0)=4.85 V

The voltage across resistor 5 is

V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V

And so the current is

I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A

The current through resistor 4 is

I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A

And therefore its voltage is

V_4=I_4 R_4 = (0.385)(2.0)=0.77 V

So, the voltage through resistor 3 is

V_3=V_5-V_4=1.17-0.77=0.4 V

And the current is

I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A

Finally, the current through resistor 2 is

I_2=I_4-I_3=0.5-0.385=0.015 A

And so its voltage is

V_2=I_2R_2=(0.015)(5.0)=0.075 V

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