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Xelga [282]
3 years ago
11

Describe what is happening to the speed during the period (I). 0s - 10s __________________________________________________ (II).

10s - 25s ________________________________________________ (III). 25 - 30 _________________________________________________
Physics
1 answer:
aleksley [76]3 years ago
4 0

Answer:

- There was a constant acceleration at 0 to 10s

- There was a zero acceleration at 10 to 25s

- There was a constant deceleration at 25 to 30s

Explanation:

<em>See attachment for complete question.</em>

Solving (a): What happens at 0s to 10s

There was a constant acceleration and this is proven below.

At time 0, velocity = 15

At time 10, velocity = 30

This is represented as:

(t_1,v_1) = (0,15)

(t_2,v_2) = (10,30)

Acceleration (A) is the rate of change of velocity against time.

So:

A = \frac{v_2 - v_1}{t_2-t_1}

A = \frac{30-15}{10 - 0}

A = \frac{15}{10}

A = 1.5

<em>Since the acceleration is positive, then it shows a constant acceleration.</em>

Solving (b): What happens at 10s to 25s

There was a zero acceleration and this is because the velocity do not change.

See proof below

At time 10, velocity = 30

At time 25, velocity = 30

This is represented as:

(t_1,v_1) = (10,30)

(t_2,v_2) = (25,30)

Acceleration (A) is the rate of change of velocity against time.

So:

A = \frac{v_2 - v_1}{t_2-t_1}

A = \frac{30-30}{25 - 10}

A = \frac{0}{15}

A = 0

Solving (c): What happens at 25s to 30s

There was a constant deceleration and this is proven below.

At time 25, velocity = 30

At time 30, velocity = 0

This is represented as:

(t_1,v_1) = (25,30)

(t_2,v_2) = (30,0)

Acceleration (A) is the rate of change of velocity against time.

So:

A = \frac{v_2 - v_1}{t_2-t_1}

A = \frac{0-30}{30-25}

A = \frac{-30}{5}

A = -6

<em>Since the acceleration is negative, then it shows a constant deceleration</em>

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Answer:

X-Positions:                                         Y-Positions

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  • It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

  • We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
  • As both axes are  perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
  • At any moment, it is subject to the acceleration of gravity, g.
  • As the acceleration is constant, we can find the vertical displacement (taking the  height of the cliff as the initial reference level), using the following kinematic equation:

       y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}

  • Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
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  • y(8) = 32*9.8 m/s2 = 313.6 m
  • y(10)= 50 * 9.8 m/s2 = 490.0 m
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