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