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A particle is moving on the real line, and its position is observed at four different time stamps:

- At time [tex]\( t = 0 \)[/tex] seconds, the particle is at [tex]\( x = 0 \)[/tex].
- At time [tex]\( t = 20 \)[/tex] seconds, the particle is at [tex]\( x = 40 \)[/tex].
- At time [tex]\( t = 40 \)[/tex] seconds, the particle is at [tex]\( x = 60 \)[/tex].
- At time [tex]\( t = 60 \)[/tex] seconds, the particle is at [tex]\( x = 90 \)[/tex].

Show that at some point in time between 0 and 60 seconds, the acceleration of the particle was zero.



Answer :

GinnyAnswer
To show that at some point between 0 and 60 seconds the acceleration of the particle was zero, let's break down the problem using the data provided.

### Step 1: Calculate the velocity at each time interval
Velocity is defined as the change in position over the change in time. We can calculate the velocity for each interval.

1. From [tex]\( t = 0 \)[/tex] to [tex]\( t = 20 \)[/tex] seconds:
[tex]\[ v_{01} = \frac{x_1 - x_0}{t_1 - t_0} = \frac{40 - 0}{20 - 0} = 2 \text{ m/s} \][/tex]

2. From [tex]\( t = 20 \)[/tex] to [tex]\( t = 40 \)[/tex] seconds:
[tex]\[ v_{12} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{60 - 40}{40 - 20} = 1 \text{ m/s} \][/tex]

3. From [tex]\( t = 40 \)[/tex] to [tex]\( t = 60 \)[/tex] seconds:
[tex]\[ v_{23} = \frac{x_3 - x_2}{t_3 - t_2} = \frac{90 - 60}{60 - 40} = 1.5 \text{ m/s} \][/tex]

### Step 2: Calculate the acceleration at each time interval
Acceleration is defined as the change in velocity over the change in time.

1. From [tex]\( t = 20 \)[/tex] to [tex]\( t = 40 \)[/tex] seconds:
[tex]\[ a_{01} = \frac{v_{12} - v_{01}}{t_2 - t_1} = \frac{1 - 2}{40 - 20} = \frac{-1}{20} = -0.05 \text{ m/s}^2 \][/tex]

2. From [tex]\( t = 40 \)[/tex] to [tex]\( t = 60 \)[/tex] seconds:
[tex]\[ a_{12} = \frac{v_{23} - v_{12}}{t_3 - t_2} = \frac{1.5 - 1}{60 - 40} = \frac{0.5}{20} = 0.025 \text{ m/s}^2 \][/tex]

### Step 3: Analyzing the acceleration values
We have found that:
- The acceleration from [tex]\( t = 20 \)[/tex] to [tex]\( t = 40 \)[/tex] seconds is [tex]\( -0.05 \text{ m/s}^2 \)[/tex].
- The acceleration from [tex]\( t = 40 \)[/tex] to [tex]\( t = 60 \)[/tex] seconds is [tex]\( 0.025 \text{ m/s}^2 \)[/tex].

Since the acceleration changes sign (from negative to positive) between these intervals, there must be a point where the acceleration is zero due to the Intermediate Value Theorem (IVT). The IVT states that if a continuous function changes sign over an interval, then there must be some point within that interval where the function is zero.

Therefore, between [tex]\( t = 20 \)[/tex] seconds and [tex]\( t = 60 \)[/tex] seconds, there exists at least one moment where the particle's acceleration is zero. This demonstrates that at some point between 0 and 60 seconds, the particle had zero acceleration.

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