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Finger [1]
3 years ago
8

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

at 1.2 m/s, while you like a brisker pace of 1.6 m/s. Knowing this, you pick a restaurant between the two buildings at which you and your friend will arrive at the same instant if both of you leave your respective buildings at the same instant. In blocks, how far from the wife’s building is the restaurant?
Physics
1 answer:
Pavel [41]3 years ago
7 0

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

s_1 = v_1t = 1.6t

Similarly the distance (m) from your friend building to the restaurant:

s_2 = v_2t = 1.2t

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

s_1 + s_2 = 5b

1.6t + 1.2t = 5b

2.8t = 5b

t = 5b/2.8 = 25b/14

s_2 = 1.2t = 1.2(25b/14) = 2.143b

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

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Please help! Will mark best answer!!
MA_775_DIABLO [31]
The answer is (A) hope it helps 
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7 0
4 years ago
A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di
navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

X(van)=5.65t+154

X(driver)=34.4t+\frac{(-2)t^{2} }{2}

or by rearanging the drivers equation.

X(driver)=34.4t+t^{2}

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

X(van)=X(driver)

5.65t+154=34.4t-t^{2}

0=t^{2} -(34.4-5.65)t+1540=t^{2} -28.75t+154

To solve this equation we use the following formulas

t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}

t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}

Where a=1; b=-28.75; c=154

So we get:

t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63st=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

V(driver)=V_{0} +at}

V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck

8 0
3 years ago
¿CUAL ES LA CAUSA DE UNA ONDA ESTACIONARIA?
xxTIMURxx [149]
Fenomeno de resonancia.

Espero que estoy te ayude :)
6 0
3 years ago
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