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4 years ago

5

On a road trip, a driver achieved an average speed of (48.0+A) km/h for the first 86.0 km and an average speed of (43.0-B) km/h

for the remaining 54.0 km. What was her average speed (in km/h) for the entire trip? Round your final answer to three significant figures.

1 answer:

tensa zangetsu [6.8K]4 years ago

6 0

Answer:

45.94 km/h

Explanation:

Let the average speed be 48km/h for the first 86 km and 43 km/h for the remaining 54 km. We can calculate the time it takes to travel each of the segment:

For the first 86 km segment:

$t_1 = s_1/v_1 = 86 / 48 = 1.79 hours$

For the remaining 54 km segment:

$t_2 = s_2/v_2 = 54 / 43 = 1.256 hours$

So the total times it takes is

$t = t_1 + t_2 = 1.79 + 1.256 = 3.047 hours$

And the total distance she travels:

$s = s_1 + s_2 = 86 + 54 = 140 km$

The average speed for the entire trip would be total distance divided by total time:

$v = s / t = 140 / 3.047 = 45.94 km/s$

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Answer:

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$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

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Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

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$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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