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GREYUIT [131]
4 years ago
13

You are driving at 35 m/s east and notice another car that is initially located 462 m in front of you and is moving east at 25 m

/s. How far do you travel before you pass the other car?
Physics
2 answers:
GaryK [48]4 years ago
7 0

The easiest way to answer this question is by realizing there are relating the velocities of the two cars. To tackle this problem, you have to understand the picture.  Car 1 travels at 35m/s and Car 2 travels at 25m/s.  Based on relative velocities, we can understand that Car 1 travels 10m/s faster than Car 2 every second.  So we can interpret Car 1's relative velocity to Car 2 as 10m/s.  Car 1 needs to travel 10m/s till a point of catching up to Car 2 which is 462m away.

v = 10m/s

d = 462m

v = d/t

(10) = (462)/t

t = 46.2s

So it takes 46.2 seconds for Car 1 to catch up to Car 2, but the question is asking how far does Car 1 travel to catch up.  So we have to use Car 1's velocity and not the relative velocity:

v = 35m/s

v = d/t

(35) = d/(46.2)

d = 1617m

Car 1 traveled a total distance of 1617m.

Bad White [126]4 years ago
3 0

-- You are 462 meters behind.

-- You're gaining on him at the rate of  (35m/s - 25m/s)= 10 m/s.

-- At that rate, it'll take you (462m / 10m/s)= 46.2 seconds to over take him.

-- During that time, you'll cover (46.2s x 35m/s)= <em>1,617 meters.</em>

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Explanation:

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Field due to charge = 4 X 10⁻⁹ C

E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}

These two fields will be equal and opposite to make net field zero

\frac{9\times 10^9\times3\times10^{-9}}{d^2} = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

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Explanation:

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