The velocity of a particle defined by the relation v = 8 -0.02x, where v is expressed in m/s and x in meter. Knowing that x = 0 at t = 0, the acceleration at t=0 is

Options

A. `-0.16 m//s^(2)`
B. `-0.02 m//s^(2)`
C. `7.08 m//s^(2)`
D. `8 m//s^(2)`

Answer

To find the acceleration at \( t = 0 \) for the given velocity function \( v = 8 - 0.02x \), we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and displacement.
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, which can be expressed as:
\[ a = \frac{dv}{dt} \]
However, we can use the chain rule to express this in terms of displacement \( x \):
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is the velocity \( v \).

### Step 2: Differentiate the velocity function with respect to \( x \).
Given the velocity function:
\[ v = 8 - 0.02x \]
we differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = -0.02 \]

### Step 3: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula.
From the chain rule, we can substitute:
\[ a = \frac{dv}{dx} \cdot v \]
Substituting \( \frac{dv}{dx} = -0.02 \) and \( v = 8 - 0.02x \):
\[ a = (-0.02) \cdot (8 - 0.02x) \]

### Step 4: Evaluate at \( x = 0 \).
Since we know that \( x = 0 \) at \( t = 0 \):
\[ a = (-0.02) \cdot (8 - 0.02 \cdot 0) = (-0.02) \cdot 8 = -0.16 \, \text{m/s}^2 \]

### Final Answer:
The acceleration at \( t = 0 \) is:
\[ a = -0.16 \, \text{m/s}^2 \]

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