A ball starts from rest and rolls down a hill with uniform acceleration, traveling 170 m during the second 5.5 s of its motion.

How far did it roll during the first 5.5 s of motion?
Express your answer using two significant figures.

x= m

Speed at middle of second 5.5 s interval = 3 x (Speed at middle of first 5.5 s interval) = 170/5.5 = 30.91 m/s

= (average speed during that interval)

Speed at 8.25 s = 3 x speed at 2.75 s.
Therefore:
speed at 2.75 s = (1/3)*30.91 = 10.30 m/s
Distance travelled during first interval 10.30 x 5.5 = 56.65 m
Rounding that two 2 significant figures gives 57 m.

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To find out how far the ball rolled during the first 5.5 seconds, we need to use the equation of motion for uniformly accelerated motion:

x = ut + (1/2)at^2

Where:
x = distance
u = initial velocity
a = acceleration
t = time

In this case, the ball starts from rest, so the initial velocity (u) is 0. The acceleration can be calculated using the distance traveled during the second 5.5 seconds:

170 m = (1/2)a(5.5 s)^2

Simplifying the equation:

170 m = (1/2)a(30.25 s^2)

Now, we can solve for acceleration (a):

a = (2 * 170 m) / (30.25 s^2)
a ≈ 11.29 m/s^2

Now that we know the acceleration, we can use the same equation to find the distance traveled during the first 5.5 seconds:

x = (1/2)(11.29 m/s^2)(5.5 s)^2

Simplifying the equation:

x = (1/2)(11.29 m/s^2)(30.25 s^2)

x ≈ 97.34 m

Therefore, the ball rolled approximately 97.34 meters during the first 5.5 seconds of its motion.

To find the distance the ball rolled during the first 5.5 seconds of motion, we can use the equation of motion:

x = ut + (1/2)at^2

Where:
x is the distance traveled,
u is the initial velocity,
t is the time elapsed, and
a is the acceleration.

In this case, the initial velocity is zero since the ball starts from rest. Therefore, the equation simplifies to:

x = (1/2)at^2

The total distance traveled during the second 5.5 seconds of motion is given as 170 m. Therefore, we can find the acceleration using this distance.

170 = (1/2)a(5.5^2)

Simplifying this equation, we have:

a = (2 * 170) / (5.5^2)

Evaluating this expression, we get:

a ≈ 10.78 m/s^2

Now that we have the acceleration, we can use it to find the distance traveled during the first 5.5 seconds of motion.

x = (1/2)at^2
x = (1/2)(10.78)(5.5^2)

Evaluating this expression, we get:

x ≈ 82.745 m

Therefore, the ball rolled approximately 82.745 m during the first 5.5 s of motion.