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A body starts from rest and is uniformly accelerated for 30 s. The distance traveled in the first 10 s is x, the next 10 s is y, and the last 10 s is z. Then x:y:z is the same as: __________

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Final answer:

The body travels a distance of 50a in the first 10 seconds, another 50a in the next 10 seconds, and 5at in the last 10 seconds.

Step-by-step explanation:

Given that the body starts from rest and is uniformly accelerating for 30 s, we can find the distance traveled in each 10 s interval.

In the first 10 s, the body starts from rest, so its initial velocity is 0 m/s. Using the equation:

d = Vit + 0.5at^2

We can substitute the values for the first 10 s interval:

x = (0)(10) + 0.5(a)(10^2)

x = 0 + 0.5(a)(100)

x = 50a

In the next 10 s, the body still starts from rest, so its initial velocity is still 0 m/s. Using the same equation, we can substitute the values for the second 10 s interval:

y = (0)(10) + 0.5(a)(10^2)

y = 0 + 0.5(a)(100)

y = 50a

In the last 10 s, the body is still uniformly accelerating, but now its initial velocity is not 0 m/s. We can use the equation:

d = Vit + 0.5at^2

Since the initial velocity is not 0 m/s, we need to find the value of the initial velocity. Using the equation:

v = u + at

Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given that the body starts from rest, the initial velocity is 0 m/s. So the equation simplifies to:

v = 0 + at

Solving for a, we have:

a = v/t

Substituting the values for the third 10 s interval:

z = (0)t + 0.5(a)(t^2)

z = 0 + 0.5(at)(10)

z = 5at

To find the ratio x:y:z, we divide the distances by a:

x:y:z = x/a:y/a:z/a

x:y:z = 50a/a:50a/a:5at/a

x:y:z = 50:50:5t

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