The velocity of a particle moving with constant acceleration at an instant `t_(0)` is `10m//s` After 5 seconds of that instant the velocity of the particle is 20m/s. The velocity at 3 second before `t_(0)` is:-

Options

A. `1/2`
B. `1`
C. `2`
D. `4`

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration.

### Step 1: Define the variables
- Let \( v_0 \) be the velocity at time \( t_0 - 3 \) seconds (the time we want to find).
- At \( t_0 \), the velocity \( v(t_0) = 10 \, \text{m/s} \).
- At \( t_0 + 5 \) seconds, the velocity \( v(t_0 + 5) = 20 \, \text{m/s} \).
- Let \( u \) be the initial velocity at \( t = 0 \).
- Let \( a \) be the constant acceleration.

### Step 2: Write the equations of motion
Using the first equation of motion:
1. At \( t_0 \):
\[ v(t_0) = u + a(t_0) \quad \text{(Equation 1)} \]
Given \( v(t_0) = 10 \, \text{m/s} \), we have:
\[ 10 = u + a(t_0) \quad \text{(1)} \]

2. At \( t_0 + 5 \):
\[ v(t_0 + 5) = u + a(t_0 + 5) \quad \text{(Equation 2)} \]
Given \( v(t_0 + 5) = 20 \, \text{m/s} \), we have:
\[ 20 = u + a(t_0 + 5) \quad \text{(2)} \]

### Step 3: Subtract the equations
To eliminate \( u \), we can subtract Equation 1 from Equation 2:
\[ 20 - 10 = [u + a(t_0 + 5)] - [u + a(t_0)] \]
This simplifies to:
\[ 10 = a(t_0 + 5) - a(t_0) \]
\[ 10 = 5a \]
From this, we can solve for \( a \):
\[ a = \frac{10}{5} = 2 \, \text{m/s}^2 \]

### Step 4: Find the velocity at \( t_0 - 3 \)
Now, we need to find the velocity at \( t_0 - 3 \):
Using the first equation of motion again:
\[ v(t_0 - 3) = u + a(t_0 - 3) \quad \text{(Equation 3)} \]
From Equation 1, we can express \( u \):
\[ u = 10 - a(t_0) \]
Substituting \( a = 2 \, \text{m/s}^2 \):
\[ u = 10 - 2(t_0) \]
Now substituting \( u \) into Equation 3:
\[ v(t_0 - 3) = (10 - 2(t_0)) + 2(t_0 - 3) \]
Simplifying this:
\[ v(t_0 - 3) = 10 - 2(t_0) + 2t_0 - 6 \]
\[ v(t_0 - 3) = 10 - 6 = 4 \, \text{m/s} \]

### Final Answer
The velocity at 3 seconds before \( t_0 \) is \( 4 \, \text{m/s} \).

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