The velocity v of a body moving along a straight line varies with time t as `v=2t^(2)e^(-t)`, where v is in m/s and t is in second. The acceleration of body is zero at t =

Options

A. 0
B. 2 s
C. 3
D. Both (1) & (2)

Answer

To find the time \( t \) at which the acceleration of the body is zero, we start with the given velocity function:

\[ v(t) = 2t^2 e^{-t} \]

### Step 1: Find the acceleration
Acceleration \( a(t) \) is the rate of change of velocity with respect to time, which is given by:

\[ a(t) = \frac{dv}{dt} \]

### Step 2: Differentiate the velocity function
We will use the product rule to differentiate \( v(t) \). The product rule states that if you have two functions \( u(t) \) and \( w(t) \), then:

\[ \frac{d(uw)}{dt} = u \frac{dw}{dt} + w \frac{du}{dt} \]

In our case, let:
- \( u(t) = 2t^2 \)
- \( w(t) = e^{-t} \)

Now we differentiate both parts:
- \( \frac{du}{dt} = 4t \)
- \( \frac{dw}{dt} = -e^{-t} \)

Applying the product rule:

\[ \frac{dv}{dt} = u \frac{dw}{dt} + w \frac{du}{dt} \]

Substituting the values:

\[ \frac{dv}{dt} = 2t^2 \cdot (-e^{-t}) + e^{-t} \cdot (4t) \]

This simplifies to:

\[ \frac{dv}{dt} = -2t^2 e^{-t} + 4t e^{-t} \]

### Step 3: Factor out common terms
We can factor out \( e^{-t} \):

\[ \frac{dv}{dt} = e^{-t}(-2t^2 + 4t) \]

### Step 4: Set acceleration to zero
To find when the acceleration is zero, we set \( \frac{dv}{dt} = 0 \):

\[ e^{-t}(-2t^2 + 4t) = 0 \]

Since \( e^{-t} \) is never zero, we focus on the quadratic part:

\[ -2t^2 + 4t = 0 \]

### Step 5: Solve the quadratic equation
Factoring out \( -2t \):

\[ -2t(t - 2) = 0 \]

This gives us two solutions:

\[ t = 0 \quad \text{or} \quad t = 2 \]

### Conclusion
The acceleration of the body is zero at \( t = 0 \) seconds and \( t = 2 \) seconds.