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DiKsa [7]
4 years ago
13

How do you calculate relative velocity?

Physics
1 answer:
nekit [7.7K]4 years ago
4 0

Answer:

<h3>Vab=Va-Vb</h3><h3>Vab=Vb-Va</h3><h3>Explanation:</h3>

Vab is velocity of a relative to b

Vba is velocity of b relative to a

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Answer:

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7 0
3 years ago
Read 2 more answers
An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

So we have v_{AT}=(95km/h)-(75Km/h)=20Km/h.

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s

And in that time the car would have traveled (<em>relative to the ground</em>):

d=v_At=(95Km/h)(0.065h)=6.175Km

If they are traveling in opposite directions, <u>we have to do all the same</u> but using v_{AT}=v_A+v_T (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

v_{AT}=(95km/h)+(75Km/h)=170Km/h

t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

5 0
3 years ago
A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc
Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity v_1  in time t_1

\frac{d}{2}=v_1t_1----1

from B to C

distance traveled is 0.5 d with v_2  and v_3  velocity for half-half time

\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2

divide 1 and 2 we get

\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}

\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}

Now average velocity is given by

v_{avg}=\frac{d}{t_1+t_2}

taking t_1  common

v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}

v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}

v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}  

6 0
3 years ago
To determine the location of the center of mass (or center of gravity) of a car, the car is drivenover a scale on a horizontal f
just olya [345]

Explanation:

It is given that,

When the front wheels are over the scale, the weight recorded by the scale is 5800 N, F₁ = 5800 N

When the rear wheels are over the scale, the scale reads 6500 N, F₂ = 6500

The distance between the front and rear wheels is measured to be 3.20 m, x₂ = 3.2 m

We need to find the location of center of mass behind the front wheels. Let the center of is located at a distance of x₁. Thus balancing the torques we get :

5800\times x_2=6500\times (3.2-x_2)

On solving the above equation we get, x₂ = 1.69 m

So, the center of mass is located at a distance of 1.69 meters behind the front wheels.

4 0
3 years ago
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