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A car travels in the + x-direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is vav-x = 6.25 m/s. How far does the car travel in 4.00 s?

Respuesta :

asuwafohjames

Answer:

25 m

Explanation:

The relationship between Velocity, distance and time is given as

S = v/t........................... Equation 1

Where S = average velocity of the car, d = distance covered by the car, t = time

Making d the subject of the equation,

d = vt.................... Equation 2

Given: v = 6.25 m/s, t = 4.00 s.

Substitute into equation 2,

d = 6.25(4)

d = 25 m.

Hence, the distance traveled by the car = 25 m

Cricetus

The distance traveled by the car will be "25 m".

The given values are:

Speed,

  • v = 6.25 m/s,

Time,

  • t = 4.00 s

As we know the formula,

→ [tex]Distance = Speed\times Time[/tex]

or,

→ [tex]d = v\times t[/tex]

By substituting the values, we get

     [tex]= 6.25\times 4[/tex]

     [tex]=25 \ m[/tex]  

Thus the above answer is right.  

Learn more about velocity here:

https://brainly.com/question/13213887

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