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3 years ago

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Ilya and Anya each can run at a speed of 7.70 mph and walk at a speed of 4.00 mph. They set off together on a route of length 5.

00 miles. Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
A) How long does it take Anya to cover the distance of 5.00 miles?
B) Find Anya's average speed.
C) How long does it take Ilya to cover the distance?
D) Now find Ilya's average speed

1 answer:

Oliga [24]3 years ago

4 0

Answer:

Explanation:

Llya and anya run speed is 7.7mph

Anya and play walk speed is 4mph

They want to travel a total distance of 5miles

Anya walks half of the journey and run half of the journey

Llya walks half of the time and runs half of the time.

A. How long does it take Anya to cover the distance of 5.00 miles?

Since anya walk half which is 2.5miles

So, she runs half which is 2.5 miles

Speed is given as distance /time

S=d/t

Therefore walking time us

t=d/S., d=2.5mile and S=4mph

t=2.5/4

t=0.625h

The running time

t=d/S. d =2.5mile and S=7.7mph

t=2.5/7

t=0.357h

Total time is 0.625+0.357

t=1.007hour

B. Average speed of anya

Average speed is given by total distance travel over total time traveled

A.S=total distance/time

Total distance is 5mile and time =1.007h

A.S=5/1.007

A.S=4.96mph

A.s=4.96miles/hour

C. How long does it take Ilya to cover the distance?

Let the total time llya use to travel be t

Then walk time =0.5t

Running time =0.5t

Total distance travel is 5miles

Total distance = speed ran × time ran + speed walked × time walked

5=7.7×0.5t+4×0.5t

5=3.85t+2t

5=5.85t

Then, t=5/5.85

t=0.855h.

D. Llya average speed

Average speed is given by total distance travel over total time traveled

A.S=total distance/total time

A.S=5/0.855

A.S=5.85mph

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a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

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e) The final speed of the system is 3.551 meters per second.

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a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

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$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

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$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

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c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

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$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

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If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

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$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

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