A car increases uniformly from rest to a rate of 21.0 km/h in 5.6 s. Discover the street the car travels during this time. Price in devices of m.
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You are watching: A car accelerates uniformly from rest to a speed of


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The motion of the car has a continuous acceleration. In this situation we will usage the expression that the acceleration and the expression of distance for the uniformly varied movement.

The acceleration is characterized as:

a = (v – v0)/t

Where:

v – v0 → sport of speed, at time...


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The activity of the car has a continuous acceleration. In this situation we will use the expression that the acceleration and also the expression of distance for the uniformly differed movement.

The acceleration is defined as:

a = (v – v0)/t

Where:

 v – v0 → sport of speed, in ~ time t

The rate variation, is from zero to 21 km/h in a time the 5.6 seconds.

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21 km/h = (21)(1000/3600) = 5.83 m/s

Then the acceleration is:

a = (5.83 – 0)/5.6 = 1.04 m/s^2

Now, we apply the expression of distance:

d = v0t + at^2/2

d = 0 + (1.04)(5.6)^2/2

d = 16.3 m

Then, this car travels 16.3 m in a time the 5.6 s


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