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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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