A body travels 2m in the first two second and 2.20m in the next 4 second with uniform deceleration. The velocity of the body at the end of 9 second is

Options

A. `-10 ms^(-1)`
B. `-0.20ms^(-1)`
C. `-0.40ms^(-1)`
D. `-0.80 ms^(-1)`

Answer

To solve the problem step by step, we need to analyze the motion of the body under uniform deceleration. We are given two displacements over specific time intervals and need to find the velocity at the end of 9 seconds.

### Step 1: Analyze the first interval (0 to 2 seconds)
The body travels 2 meters in the first 2 seconds. We can use the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
where:
- \( S \) is the displacement (2 m),
- \( u \) is the initial velocity,
- \( t \) is the time (2 s),
- \( a \) is the acceleration.

Substituting the known values:
\[ 2 = u(2) + \frac{1}{2} a (2^2) \]
\[ 2 = 2u + 2a \]
Dividing the entire equation by 2:
\[ 1 = u + a \quad \text{(Equation 1)} \]

### Step 2: Analyze the second interval (2 to 6 seconds)
In the next 4 seconds (from 2 to 6 seconds), the body travels 2.2 meters. The initial velocity at \( t = 2 \) seconds is given by:
\[ v = u + at \]
At \( t = 2 \):
\[ v = u + 2a \]

Now, we can use the equation of motion again for the interval from 2 to 6 seconds:
\[ S = ut + \frac{1}{2} a t^2 \]
Here, \( S = 2.2 \), \( t = 4 \) (from 2 to 6 seconds), and the initial velocity is \( v = u + 2a \):
\[ 2.2 = (u + 2a)(4) + \frac{1}{2} a (4^2) \]
\[ 2.2 = 4u + 8a + 8a \]
\[ 2.2 = 4u + 16a \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( u + a = 1 \)
2. \( 4u + 16a = 2.2 \)

From Equation 1, we can express \( u \) in terms of \( a \):
\[ u = 1 - a \]

Substituting this into Equation 2:
\[ 4(1 - a) + 16a = 2.2 \]
\[ 4 - 4a + 16a = 2.2 \]
\[ 4 + 12a = 2.2 \]
\[ 12a = 2.2 - 4 \]
\[ 12a = -1.8 \]
\[ a = -0.15 \, \text{m/s}^2 \]

Now substituting \( a \) back into Equation 1 to find \( u \):
\[ u + (-0.15) = 1 \]
\[ u = 1 + 0.15 \]
\[ u = 1.15 \, \text{m/s} \]

### Step 4: Find the velocity at the end of 9 seconds
Now we can find the velocity at \( t = 9 \) seconds using the formula:
\[ v = u + at \]
Substituting the values:
\[ v = 1.15 + (-0.15)(9) \]
\[ v = 1.15 - 1.35 \]
\[ v = -0.20 \, \text{m/s} \]

### Final Answer
The velocity of the body at the end of 9 seconds is \( -0.20 \, \text{m/s} \).

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