Question

Describe what is happening to the speed during the period (I). 0s - 10s __________________________________________________ (II). 10s - 25s ________________________________________________ (III). 25 - 30 _________________________________________________

Answer 1

- 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

Step-by-step explanation:

See attachment for complete question.

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 = (v_2 - v_1)/(t_2-t_1)`

`A = (30-15)/(10 - 0)`

`A = (15)/(10)`

`A = 1.5`

Since the acceleration is positive, then it shows a constant acceleration.

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 = (v_2 - v_1)/(t_2-t_1)`

`A = (30-30)/(25 - 10)`

`A = (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 = (v_2 - v_1)/(t_2-t_1)`

`A = (0-30)/(30-25)`

`A = (-30)/(5)`

`A = -6`

Since the acceleration is negative, then it shows a constant deceleration