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3 years ago

5

After the diver jumps forward from the diving board the force of gravity will accelerate the diver _ to the direction of motion

1 answer:

joja [24]3 years ago

5 0

I’m unsure whether there is a question there but in forces the force weight which includes gravity would be pushing the diver directly down. However if there was enough thrust from the diver the motion would change but ultimately will be accelerating downwards because of unbalanced forces

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

aleksley [76]

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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